Real Analysis

Winter 2025

Part 1:

Sets

(See Chapter 1 in Book of Proof)

The central objects in mathematics are sets. A set is a collection of things. The things in a set are called elements.

Sets

Often, we denote a set by simply writing all the elements between \(\{\}\) ("curly braces") separated by commas. For example,

\[\{1,2,4,5\}\]

is the set with four elements, these elements being \(1,2,4,\) and \(5\).

We often give sets names, these will frequently be a single capital roman letter, for example, \[B = \{1,2,4,5\}.\]

To express the statement "\(x\) is an element of \(A\)" we write \(x\in A\). Thus, given the set \(B\) defined above, we see that \(1\in B\) is a true statement, and \(3\in B\) is a false statement.

We will also write \(x\notin A\) to for the statement "\(x\) is not an element of \(A\)."

Equality of sets

Two sets \(A\) and \(B\) are equal (they are the same set) if they contain exactly the same elements, that is, the following two statements hold:

If \(x\in A\), then \(x\in B\).

If \(x\in B\), then \(x\in A\).

In this case we write \(A=B\).

Important example:

Important example: Consider the sets \(A=\{1,2,2,3\}\) and \(B=\{1,2,3\}\). We can easily see that every element of \(A\) is an element of \(B\), and every element of \(B\) is an element of \(A\). Thus

\[\{1,2,2,3\}=\{1,2,3\}.\]

An element is either in a set or it is not. Sets cannot contain multiple copies of an element.

Subsets

Given two sets \(A\) and \(B\), if every element of \(A\) is also an element of \(B\), then we say that \(A\) is a subset of \(B\), and write \(A\subset B\) or \(A\subseteq B\) (these both mean the exact same thing). For example,

\[\{1,2\}\subset\{1,2,4,5\}\]

but \(\{1,2,3\}\) is not a subset of \(\{1,2,4,5\}\) since \(3\in\{1,2,3\}\) and \(3\notin\{1,2,4,5\}\)

Note that the statement "\(A=B\)" can be rewritten as "\(A\subset B\) and \(B\subset A.\)" This will be very useful later.

Ellipsis notation

The set of integers is denoted \(\mathbb{Z}\). The set of positive integers, also called the natural numbers or the naturals, is denoted \(\mathbb{N}\) or \(\mathbb{Z}^{+}\). 

We will sometimes write sets like \(\mathbb{N}\) and \(\mathbb{Z}\) with ellipsis notation, as follows,

\[\N=\{1,2,3,\ldots\}\qquad \Z = \{\ldots,-2,-1,0,1,2,3,\ldots\}\]

In this notation, we list the first several elements of a set, until there is an apparent pattern, then write an ellipsis (...). After the ellipsis we write the final element in the pattern. If no element is written after the ellipsis, then the list of elements goes on without end.

Examples.

• \(\{2,4,6,8,\ldots\}\) is the set of even positive integers.
• \(\{3,7,11,15,\ldots,43\}=\{3,5,7,11,15,19,23,27,31,35,39,43\}\)

Problems with ellipsis notation

Consider the set

\[C:=\{2,4,8,\ldots\}.\]

Questions:

Is \(16\in C\)?    Is \(14\in C\)?

What is the pattern? Is the \(n\)th term in the sequence \(2^{n}\) or \(n^2-n+2\) or something else?

It's not clear!

Suppose \(N\in\mathbb{N}\) and define the set

\[D:=\{2,4,6,\ldots,2N\}.\] If \(N=5\), then

\[D = \{2,4,6,\ldots,10\}=\{2,4,6,8,10\}\]

but if \(N=2\), then

\[D = \{2,4,6,\ldots,4\}(?)\]

We will interpret this set as \(\{2,4\}\). 

Set builder notation

Let \(A\) be a set, and \(S(x)\) denote a statement whose truth value depends on \(x\in A\). The set of \(x\in A\) such that \(S(x)\) is true is denoted

\[\{x\in A : S(x)\}\quad\text{or}\quad \{x\in A\mid  S(x)\}\quad\text{or}\quad \{x : x\in A\text{ and }S(x)\}.\]

Both the colon and the vertical line are read as "such that" or "with the property that."

Example. Consider the set of even positive integers:

\[\{x\in\mathbb{N} : x\text{ is even}\}.\]

This would be read:

"The set of \(x\) in the natural numbers such that \(x\) is even."

Using the notation from the above definition we have \(A=\N\) and \(S(x)\) is the statement "\(x\) is even"

Set builder notation

More generally, let \(A\) be a set. For each \(x\in A\) let \(f(x)\) be an expression that depends on \(x\) and \(S(x)\) denote a statement whose truth value depends on \(x\in A\). The set of all elements of the form \(f(x)\) with the property that \(x\in A\) and \(S(x)\) is true is denoted

\[\{f(x) : x\in A\text{ and }S(x)\}.\]

Example. Consider the set of even positive integers:

\[\{2x: x\in \N\}.\]

In the notation of the above definition we have  \(A=\N\), \(f(x) = 2x\), and \(S(x)\) is true for all \(x\in\N\).

Using set builder notation we can disambiguate the sets from two slides prior:

\[\{2^{n} : n\in\mathbb{N}\}\quad\text{and}\quad\{n^2-n+2 : n\in\mathbb{N}\}.\]

Examples. 

• \(\{2n-1 : n\in\N\text{ and }n\leq 5\} = \{1,3,5,7,9\}\)
• \(\{25-4n : n\in\N\text{ and }n\leq 10\text{ and } n>5\} = \{1,-3,-7,-11,-15\}\)

Consider the set 

\[\{x\in\mathbb{R} : x^{2}+1 = 0\}.\]

There does not exist a real number \(x\) such that \(x^{2}+1=0\), and hence this set does not contain any elements!

The set containing no elements is called the empty set. The empty set is denoted \(\varnothing\) or \(\{\}\).

Part 2:

Proofs and Logic

(See Chapters 2,4,5 & 6 in Book of Proof)

Proposition 1. \(\{n\in\mathbb{Z} : n+1\text{ is odd}\} = \{2n : n\in\mathbb{Z}\}.\)

Proving sets are equal.

Proof. First we will show that \(\{n\in\mathbb{Z} : n+1\text{ is odd}\} \subset \{2n : n\in\mathbb{Z}\}.\) Let \(m\in\{n\in\mathbb{Z} : n+1\text{ is odd}\}\) be arbitrary. This implies that \(m+1\) is odd, that is, there exists \(k\in\Z\) such that \(m+1=2k+1.\) Subtracting \(1\) from both sides we deduce \(m=2k\), and hence \(m\in \{2n : n\in\Z\}.\)

Next, we will show \(\{2n:n\in\mathbb{Z}\}\subset\{n\in\mathbb{Z} : n+1\text{ is odd}\}\).

Let \(p\in\{2n:n\in\mathbb{Z}\}\) be arbitrary. Then, there exists \(q\in\mathbb{Z}\) such that \(p=2q\), and hence \[p+1 = 2q+1.\] Since \(q\in\mathbb{Z}\), we conclude that \(p+1\) is odd, and hence

\(p\in\{n\in\mathbb{Z} : n+1\text{ is odd}\}.\) \(\Box\)

Definition. An integer \(n\) is even if there exists an integer \(k\) such that \(n=2k\). An integer \(n\) is odd if there exists an integer \(p\) such that \(n=2p+1\).

Proposition 2. If \(n\in\Z\) is odd, then \(n^{2}\) is odd.

Proving implications (Direct proof)

\(A : n\) is odd

\(B : n^{2}\) is odd

We wish to prove the implication \(A\Rightarrow B\). We will use a direct proof, that is, we will assume \(A\), argue directly to \(B\).

Proof. Suppose \(n\in\Z\) is odd. By definition, this means that there exists \(k\in\Z\) such that \(n=2k+1\). This implies

\[n^{2} = (2k+1)^{2} = 4k^{2}+4k+1 = 2(2k^{2}+2k)+1.\]

Since \(2k^{2}+2k\) is an integer, we conclude that \(n^{2}\) is odd. \(\Box\)

Proposition 3. Suppose \(n\in\Z\). If \(n^{2}\) is even, then \(n\) is even.

Proving implications (Contrapositive proof)

Proposition 4. Suppose \(a,b,c\in\Z\). If \(a^{2}+b^{2}=c^{2}\), then either \(a\) is even or \(b\) is even.

Proving implications (Proof by contradiction)

Proposition 4. Suppose \(a,b,c\in\Z\). If \(a^{2}+b^{2}=c^{2}\), then either \(a\) is even or \(b\) is even.

Proposition 3. Suppose \(n\in\Z\). If \(n^{2}\) is even, then \(n\) is even.

Contrapositive vs. Contradiction

Proof. Assume toward a contradiction that \(n\) is not even and \(n^{2}\) is even. This implies that \(n\) is odd, that is \(n=2k+1\) for some \(k\in\Z\). Squaring, we obtain

\[n^{2} = 4k^{2}+4k+1=2(2k^{2}+2k)+1.\]

Since \(2k^{2}+2k\) is an integer, this shows that \(n^{2}\) is odd. This contradicts the fact that \(n^{2}\) is even.\(\Box\)

Proof by contradiction: \(A\wedge(\sim\!B)\Rightarrow C\wedge(\sim\!C)\)

Contrapositive: \((\sim\!B)\Rightarrow(\sim\!A)\)

Fake proof by contradiciton: \(A\wedge(\sim\!B)\Rightarrow A\wedge(\sim\!A)\)

Moral: If your proof by contradiction concludes with \(\sim\!A\), then you should use proof by contrapositive instead!

Universal quantifier

\(A : n^{2}\) is even

\(B : n\) is even

Note that \(A\) and \(B\) below are not statements. They're truth value depends on \(n\).

These are more precisely called open sentences (or predicates or propositional functions). They should really have variables

\(A(n) : n^{2}\) is even

\(B(n) : n\) is even

Proposition 3. Suppose \(n\in\Z\). If \(n^{2}\) is even, then \(n\) is even.

Recall the proposition:

Now, Proposition 3 can be more precisely written as \[\forall\,n\in\Z,\ A(n)\Rightarrow B(n).\]

Where the symbol \(\forall\) stands for "for all" or "for every," and is called the universal quantifier.

Note that \(A(n) : n^{2}\) is even, is false unless \(n\in\Z\).

Recall that for false statements \(A\), the statement \(A\Rightarrow B\) is true for any statement \(B\). (In this case \(A\Rightarrow B\) is called a vacuously true statement.)

Thus, in the statement \(\forall\,n\in\Z,\ A(n)\Rightarrow B(n)\) the universal quantifier isn't really necessary. It would be very common to see Proposition 3 written as

Proposition 3. If \(n^{2}\) is even, then \(n\) is even.

The universal quantifier "For all \(n\in\Z\)" is there, it's just not written.

Moral: In math we often omit the universal quantifier before an implication.

Existential quantifier

To say that an open sentence \(A(x)\) is true for all \(x\) in a set \(X\) we use the universal quantifier:

\[\forall\,x\in X,\ A(x).\]

In math, we would write this using words:

"For all \(x\in X\), \(A(x)\) is true."

To say that an open sentence \(A(x)\) is true for at least one \(x\) in a set \(X\) we use the existential quantifier \[\exists\, x\in X,\ A(x).\] In words:

"There exists \(x\in X\) such that \(A(x)\) is true."

Universally quantified statements may be more common, but here is an example of a statement that is not:

"The polynomial \(x^{5}+5x^3+10\) has a real root."

Do you see that this statement is of the form \(\exists\, x\in X,\ A(x)\). What are the set \(X\) and the open sentence \(A(x)\)?

Order of quantifiers

Consider the following two statements:

\(A:\ \) For every even \(n\in\N\) there exists an odd \(m\in\N\) such that \(m>n\).

\(B:\ \) There exists odd \(m\in\N\) such that for all even \(n\in\N\), it is true that \(m>n\).

Because it sounds better in English, we will often write \(B\) as

\(B:\ \) There exists odd \(m\in\N\) such that \(m>n\) for all even \(n\in\N\).

Let \(E\) denote the set of even natrual numbers, and let \(D\) denote the set of odd natural numbers. Using symbols, we can write \(A\) and \(B\) as follows: \[A: \forall\, n\in E,\ \exists\,m\in D,\ m>n\]

\[B: \exists\,m\in D,\ \forall\, n\in E,\ m>n\]

Moral: \(A\) is true, and \(B\) is false. The order of the quantifiers matters. You must think hard to figure out the meaning of a statement, there is no shortcut!

Proving universally quantified statements

Every universally quantified statement can be converted to an implication:

\[\forall\, x\in X,\ A(x)\]

is the same as

\[(x\in X)\Rightarrow A(x)\]

Thus, to prove this statement we take an "arbitrary" element \(x\in X\), then reason that \(A(x)\) is true for that \(x\). The proof looks like

Proof. Let \(x\in X\) be arbitrary.

\[\vdots\]

\[\vdots\]

Therefore, \(A(x)\). \(\Box\)

For example, let's say we want to prove the following statement:

Proposition. For every even number \(n\in\N\), the number \(n^2+1\) is odd.

For example, let's say we want to prove the following statement:

Proposition. For every even number \(n\in\N\), the number \(n^2+1\) is odd.

Proof. Let \(n\in\N\) be an arbitrary even number. This implies that \(n=2k\) for some integer \(k\), and thus

\[n^2+1 = (2k)^{2}+1 = 4k^{2}+1 = 2(2k^{2})+1.\]

Since \(2k^2\) is an integer, we conclude that \(n^2+1\) is odd. \(\Box\)

Proving quantified statements

Theorem 1.18. For every positive real number \(x\) there exists a natural number \(n\) such that \(n>x.\)

Proposition. For every \(\varepsilon>0\) there exists \(m\in\N\) such that \(\dfrac{1}{m}<\varepsilon\).

The following is an important property of the natural numbers and the real numbers. We will prove it later, but for now we will assume it is true.

Proving quantified statements

Proposition. For every \(\varepsilon>0\) there exists \(m\in\N\) such that \(\dfrac{1}{m}<\varepsilon\).

What would it look like if we tried to prove this statement by contradiction?

Statement: \(\forall\,\varepsilon>0,\ \exists\,m\in\N,\ \dfrac{1}{m}<\varepsilon.\)

Negation: \(\exists\,\varepsilon>0,\ \forall\,m\in\N,\ \dfrac{1}{m}\geq\varepsilon.\)

In words, the negation might read:

There exists \(\varepsilon>0\) such that \(\varepsilon\leq \dfrac{1}{m}\) for all \(m\in\mathbb{N}\).

Part 3:

Properties of \(\mathbb{R}\)

(See Chapter 1 in Spivak)

The algebraic properties of the real numbers \(\mathbb{R}\):

(P1) (Associativelaw for addition) \[\forall\,a,b,c\in\mathbb{R},\ a+(b+c) = (a+b)+c.\]

(P2) (Existence of an additive idenity) \[\exists\, 0\in\mathbb{R},\ \forall a\in\mathbb{R},\ a+0=0+a=a.\]

(P3) (Existence of additive inverses) \[\forall\, a\in\mathbb{R},\ \exists -a\in\mathbb{R},\ a+(-a) = (-a)+a = 0.\]

(P4) (Commutative law for addition) \(\forall\, a,b\in\mathbb{R},\ a+b=b+a.\)

(P5) (Associative law for multiplication)  \(\forall\,a,b,c\in\mathbb{R},\ a\cdot(b\cdot c) = (a\cdot b)\cdot c.\)

(P6) (Existence of a multiplicative identity) \[\exists\, 1\in\mathbb{R}\setminus\{0\}, \forall\, a\in\mathbb{R},\ a\cdot 1 = 1\cdot a = a.\]

(P7) (Existence of multiplicative inverses) \[\forall\, a\in\mathbb{R}\setminus\{0\},\ \exists\, a^{-1}\in\mathbb{R},\ a\cdot a^{-1} = a^{-1}\cdot a = 1.\]

(P8) (Commutative law for multiplication) \(\forall\, a,b\in\mathbb{R},\ a\cdot b = b\cdot a.\)

(P9) (Distributive law) \(\forall\,a,b,c\in\mathbb{R},\ a\cdot(b+ c) = a\cdot b + a\cdot c.\)

Notation

Some consequences:

Proposition 1. The additive identity in \(\mathbb{R}\) is unique.

Proposition 3. If \(a\in\mathbb{R}\), then \(0\cdot a = 0\).

Proof. Let \(a\in\mathbb{R}\) be arbitrary. Note that

\[0\cdot a = 0\cdot a + 0\cdot a - 0\cdot a = (0+0)\cdot a - 0\cdot a = 0\cdot a - 0\cdot a = 0.\ \Box\]

Inequalities

Given real numbers \(x\) and \(y\), we write \(x>y\) (or \(y<x\)) if \(x-y\) is a positive number.

But what is a positive number?

The set of positive numbers \(P\subset\mathbb{R}\) is the unique set satisfying:

(P10) (Trichotomy law)

\(\forall\,a\in\mathbb{R}\) exactly one of the following holds: (i) \(a=0\), (ii) \(a\in P\), (iii) \(-a\in P.\)

(P11) (Closure under addition)

\[\forall\,a,b\in P,\ a+b\in P.\]

(P12) (Closure under multiplication)

\[\forall\,a,b\in P,\ a\cdot b\in P.\]

Proposition 5. \(1>0\).

Proposition 7. If \(ab > 0\), then \(a\) and \(b\) are either both positive or both negative.

Definitions: For \(a,b\in\mathbb{R}\)

   (i) \(a<b\) means \(b-a\) is positive

   (ii) \(a>b\) means \(a-b\) is positive

   (iii) \(a\leq b\) means \(a<b\) or \(a=b\)

   (iv) \(a\geq b\) means \(a>b\) or \(a=b\).

We write \(a<b<c\) to mean both \(a<b\) and \(b<c\). (Similarly for expressions such as \(a\leq b<c\).)

More on inequalities

Absolute Value

For \(a\in\mathbb{R}\) we define the absolute value of \(a\), denoted \(|a|\) by

\[|a| = \begin{cases} a & a\geq 0,\\ -a & a<0.\end{cases}\]

Theorem 1 (The triangle inequality). For all real numbers \(a\) and \(b\), we have

\[|a+b|\leq |a|+|b|\]

Proof. (On the board and in the text.)

Other properties of \(\mathbb{R}\)

There are many other familiar properties of real numbers that we know and will use.

For example, if \(x,y,z\in\mathbb{R}\), then \(x<y\) if and only if \(x+z<y+z\). That is, given a true inequality, we can add anything to both sides and we get a new true inequality.

Similarly, if \(x,y\in\mathbb{R}\) and \(c>0\), then \(x<y\) if and only if \(cx<cy\). That is, given a true inequality, we can multiply both sides by a positive number and we get a new true inequality.

Part 4:

Functions

(Read Chapter 3 in the textbook)

Definition. A function is a collection of ordered pairs of numbers with the following property: if \((a,b)\) and \((a,c)\) are both in the collection, then \(b=c\).

Defining functions

We will often write something like

\[f(x) = x^2-x+1\]

to denote the function

\[f = \{(x,x^2-x+1) : x\in\mathbb{R}\}\]

Combining functions

Let \(f\) and \(g\) be two functions.

Composing functions

Part 5:

Limits

(Read Chapters 4 & 5 in the textbook)

Limits (informally)

The function \(f\) approaches the limit \(l\) near \(a\), if we can make \(f(x)\) as close as we like to \(l\) by requiring that \(x\) be sufficiently close to, but not equal to, \(a\).

Example. Suppose \(f\) is a function such that

\[f(1) =2+1,\ f(1/2) = 2+1/2,\ f(1/3) = 2+1/3,\ f(1/4)=2+1/4, \ldots\]

in general, \(f(1/n) = 2+\frac{1}{n}\) for \(n\in\mathbb{N}\).

Example. Consider the function \(f\) whose graph is shown below.

Examples from the text

How can we prove that \(f\) approaches \(l\) near \(a\)?

How small must we take \(\delta\)?

For any \(x\) such that \(0<|x-2|<\delta\), it must be the case that

\[|S(x) - 4|<\varepsilon\]

\[\Updownarrow\]

\[|x^2-4|<\varepsilon\]

\[|x-2||x+2|<\varepsilon\]

\[\Updownarrow\]

If \(\delta\leq 1\), then the only \(x\)'s we will consider are in \((1,3)\).

Thus \(x+2\in(3,5)\), and hence \(|x+2|=x+2<5\).

Thus, if \(\delta\leq 1\) and \(\delta<\varepsilon/5\), then for any \(x\) such that \(0<|x-2|<\delta\), it should follow that \(|f(x)-4|<\varepsilon\).

Limits (formally)

Definition. The function \(f\) approaches the limit \(l\) near \(a\) means: for every \(\varepsilon>0\) there is some \(\delta>0\) such that, for all \(x\), if \(0<|x-a|<\delta\), then \(|f(x)-l|<\varepsilon\).

Proposition. If \(S(x) = x^2\), then \(S\) approaches \(4\) near \(2\).

Proof. Let \(\varepsilon>0\) be arbitrary. Set

\[\delta = \min\{1,\varepsilon/10\}.\]

Clearly \(\delta>0\). Next, let \(x\) be an arbitrary real number such that \(|x-2|<\delta\). Since \(\delta\leq 1\), we see that \(|x-2|<1\), and thus \(x\in (1,3)\), this implies           \(x+2\in(3,5)\), and hence

\[|x+2| = x+2 <5.\]

Since \(\delta\leq \varepsilon/10\) we see that \(|x-2|<\varepsilon/10\). Finally, we have

\[|S(x)-4| = |x^2-4| = |(x-2)(x+2)| = |x-2||x+2|\leq \frac{\varepsilon}{10}\cdot 5 = \frac{\varepsilon}{2}<\varepsilon.\]

Proposition. Suppose \(L(x) = 2x-3\). Then \(L\) approaches \(1\) near \(2\).

Proof. Let \(\varepsilon>0\) be arbitrary. Set \(\delta = \varepsilon/2\). Clearly \(\delta>0\). Let \(x\) be an arbitrary real number such that

\[0<|x-2|<\delta = \frac{\varepsilon}{2}.\]

Hence,

\[|L(x) - 1| = |2x-3-1| = |2x-4| =2|x-2|<2\cdot\frac{\varepsilon}{2} = \varepsilon.\ \Box\]

Proposition. Suppose \(R(x) = 1/x\). Then \(R\) approaches \(2\) near \(1/2\).

From the picture it appears that  we should take

\[\delta = \min\left(\frac{1}{2-\varepsilon} - \frac{1}{2}, \frac{1}{2} - \frac{1}{2+\varepsilon}\right)\]

\[=\min\left(\frac{\varepsilon}{2(2-\varepsilon)},\frac{\varepsilon}{2(2+\varepsilon)}\right)\]

\[=\frac{\varepsilon}{2(2+\varepsilon)}\]

Proposition. Suppose \(R(x) = 1/x\). Then \(R\) approaches \(2\) near \(1/2\).

Proof. Let \(\varepsilon>0\) be arbitrary. Set

\[\delta = \frac{\varepsilon}{2(2+\varepsilon)}.\]

Let \(x\) be an arbitrary real number such that

\[0<\left|x-\frac{1}{2}\right| <\delta =  \frac{\varepsilon}{2(2+\varepsilon)}.\]

Observe that

\[x-\frac{1}{2}>-\delta = \frac{1}{2+\varepsilon}-\frac{1}{2}.\]
This implies

\[x>\frac{1}{2+\varepsilon}.\]

From this we see that \(x>0\) and furthermore

\[\frac{1}{|x|} = \frac{1}{x}<2+\varepsilon.\]

(\(\ast\))

Proof continued. Using (\(\ast\)) we have

\[|R(x) - 2| = \left|\frac{1}{x}-2\right| = \left|\frac{1-2x}{x}\right| = 2\left|x-\frac{1}{2}\right|\frac{1}{|x|}<2\cdot \frac{\varepsilon}{2(2+\varepsilon)}\cdot (2+\varepsilon)\]

\[=\varepsilon.\]

Thus for any \(x\) satisfying \(0<|x-\frac{1}{2}|<\delta\) we see that \(|R(x)-2|<\varepsilon.\) Since \(\varepsilon>0\) was arbitrary, we conclude that \(R\) approaches \(2\) near \(1/2\). \(\Box\)

Proof take 2. Let \(\varepsilon>0\) be arbitrary. Set

\[\delta = \min\left(\frac{1}{4},\frac{\varepsilon}{16}\right).\]

Let \(x\) be an arbitrary real number such that

\[0<\left|x-\frac{1}{2}\right| <\delta.\]

Observe that since \(\delta\leq1/4\) we see that \(1/4<x<3/4\). From this we deduce that \(x>0\), and that \(1/|x| = 1/x<4\). Finally, we conclude

\[|R(x) - 2| = \left|\frac{1}{x}-2\right| = \left|\frac{1-2x}{x}\right| = 2\left|x-\frac{1}{2}\right|\frac{1}{|x|}\leq 2\cdot \frac{\varepsilon}{16}\cdot 4 = \frac{\varepsilon}{2} <\varepsilon.\]

Proposition. Suppose \(R(x) = 1/x\). Then \(R\) approaches \(2\) near \(1/2\).

Thus for any \(x\) satisfying \(0<|x-\frac{1}{2}|<\delta\) we see that \(|R(x)-2|<\varepsilon.\) Since \(\varepsilon>0\) was arbitrary, we conclude that \(R\) approaches \(2\) near \(1/2\). \(\Box\)

Theorem 1. A function cannot approach two different limits near \(a\). In other words, if \(f\) approaches \(l\) near \(a\) and \(f\) approaches \(m\) near \(a\), then \(l=m\).

Proof. (In the text and on the board.)

           (1) If \(|x-x_{0}|<\frac{\varepsilon}{2}\) and \(|y-y_{0}|<\frac{\varepsilon}{2}\), then

\[|(x+y)-(x_{0}+y_{0})|<\varepsilon.\]

If \[|x-x_{0}|<\min\left(1,\frac{\varepsilon}{2(|y_{0}|+1)}\right)\quad\text{and}\quad |y-y_{0}|<\frac{\varepsilon}{2(|x_{0}|+1)},\] then \[|xy-x_{0}y_{0}|<\varepsilon.\]

If \(y_{0}\neq 0\) and

\[|y-y_{0}|<\min\left(\frac{|y_{0}|}{2},\frac{\varepsilon|y_{0}|^{2}}{2}\right),\] then \(y\neq 0\) and

\[\left|\frac{1}{y}-\frac{1}{y_{0}}\right|<\varepsilon.\]

Proof. (In the text and on the board.)

Lemma.

(2)

(3)

Theorem 2. If

\[\lim_{x\to a}f(x) = l\quad\text{and}\quad \lim_{x\to a}g(x) = m,\]

then

\[\lim_{x\to a}(f+g)(x) = l+m\]

\[\lim_{x\to a}(f\cdot g)(x) = l\cdot m.\]

Moreover, if \(m\neq 0\), then

\[\lim_{x\to a}\left(\frac{1}{g}\right)(x) = \frac{1}{m}.\]

Proof. (In the text and on the board.)

Proposition. If \(a\) and \(c\) are real numbers, then

\[\lim_{x\to a}c = c\quad\text{and}\quad\lim_{x\to a}x = a.\]

Limits of polynomials and rational functions.

A function \(f\) is called a polynomial function if there are numbers \(a_{0},a_{1},\ldots,a_{n}\) such that

\[f(x) = a_{0}+a_{1}x+a_{2}x^2+\cdots + a_{n}x^{n}\quad\text{ for all } x.\]

A function \(g\) is called a rational function if there are two polynomial function \(p\) and \(q\) such that \(g=p/q\).

Limits from above and below

Consider the function \(\displaystyle{f(x) = \begin{cases} 2x+1 & x>1\\ 1-x^2 & x<1\\ 1 & x=1.\end{cases}}\)

Part 6:

Continuity

(Read Chapter 6 in the textbook)

The function \(f\) is continuous at \(a\) if

\[\lim_{x\to a}f(x) = f(a).\]

Continuity at a point

Theorem 1. If \(f\) and \(g\) are continuous at \(a\), then \(f+g\) and \(f\cdot g\) are continuous at \(a\). Moreover, if \(g(a)\neq 0\), then \(1/g\) is continuous at \(a\).

Continuity at on an interval

If \(f\) is continuous at every point \(c\) in the interval \((a,b)\), then we say \(f\) is continuous on \((a,b)\).

If \(f\) is continuous at every real number \(c\), then we say \(f\) is continuous on \((-\infty,\infty)\).

If \(f\) is continuous at every real number \(c\) in the interval \((a,b)\), and

\[\lim_{x\to b^{-}}f(x) = f(b)\quad\text{and} \lim_{x\to a^{+}}f(x) = f(a),\] then we say \(f\) is continuous on \([a,b]\).

We make similar definitions for continuity on \([a,b)\), \((a,b]\), \((-\infty,b)\), \((-\infty,b]\), \((a,\infty)\), and \([a,\infty)\).

Part 7:

Three Hard Theorems

(Read Chapter 7 in the textbook)

Informally, a function \(f\) is continuous on \([a,b]\) if the graph of \(f\) can be drawn without picking up your pencil, starting with your pencil at \((a,f(a))\) and ending at \((b,f(b))\).

The intuitive notion of continuity

1. If you were below sea level at some time \(t_{0}\), then you were above sea level at some later time \(t_{1}\), then there was a time \(s\), between \(t_{0}\) and \(t_{1}\) when you were exactly at sea level.
2. Over a finite period of time, your elevation was not arbitrarily large (even if you own a spaceship).
3. Over a finite period of time, there was some point (maybe more than one) where your elevation was largest.

Motivating example

Let

\(A(t) = \) your current distance above sea level,

where \(t = \)  number of seconds since you were born.

Bolzono's Theorem

The statement \(f\) is continuous on \([a,b]\) is meant to imply a precise version of the intuitive statement

"You can draw the graph without picking up your pencil."

meant to imply

Continuous on \([a,b]\ \Rightarrow\) bounded

If \(f\) is continuous on \([a,b]\), then \(f(x)\) should not get arbitrarily large for \(x\) in \([a,b]\), but does our precise definition of continuity imply this?

Extreme value theorem

If \(f\) is continuous on \([a,b]\), then \(f(x)\) should achieve it's maximum value on \([a,b]\). Does our definition imply this?

Some consequences of the Big Three:

The intermediate value theorem

Some consequences of the Big Three:

Extreme value theorem from below

Some consequences of the Big Three:

Roots and polynomials

Theorem 8. If \(\alpha>0\), then there exists a number \(x\) such that \(x^{2}=\alpha\).

What about even \(n\)?

Part 8:

Least upper bounds

(Read Chapter 8 in the textbook)

Bounded sets

Definition. A set of real numbers \(A\) is bounded above if thereis a number \(x\) such that

\[x\geq a\quad\text{for every }a\text{ in }A.\]

Such a number \(x\) is called an upper bound for \(A\).

Least upper bound

Definition. A number \(x\) is called a least upper bound of \(A\) if

(1)  \(x\) is an upper bound for \(A\), and 

(2)  if \(y\) is an upper bound for \(A\), then \(x\leq y\).

Consider the set

\[A = \{x : x^2<2\}.\]

Does this set have a least upper bound?

Least upper bound property

We finally arrive at the final property of the real numbers:

(P13)

(Least upper bound property) If \(A\) is a set of real numbers, \(A\neq\varnothing\), and \(A\) is bounded above, then \(A\) has a least upper bound.

Notation

Given a set of numbers \(A\), if \(A\) has a least upper bound, then we denote the least upper bound of \(A\) with the symbol

\[\sup A.\]

The least upper bound of \(A\) is also known as the supremum of \(A\).

We have similar definitions for a lower bound of a set, a set which is bounded below, and the greatest lower bound. The greatest lower bound of a set is denoted

\[\inf A,\]

and it is also known as the infimum of the set \(A\).

Proof of Theorem 1-7

Theorem 1 (Bolzano's Theorem). If \(f\) is continuous on \([a,b]\) and \(f(a)<0<f(b)\), then there is some \(x\) in \([a,b]\) such that \(f(x)=0\).

We're going to need the bump theorem, and some variations of it:

A lemma

Theorem 1. If \(f\) is continuous at \(a\), then there is a number \(\delta>0\) such that \(f\) is bounded above on the interval \((a-\delta,a+\delta)\).

Theorem 2. If \(f\) is continuous on \([a,b]\), then \(f\) is bounded above on \([a,b]\), that is, there is a number \(N\) such that \(f(x)\leq N\) for all \(x\) in \([a,b]\).

Let's prove another.

Proof idea. Set

\[A = \{x : a\leq x\leq b,\ f\text{ is bounded above on }[a,x]\},\]

and set

\[ c = \sup A.\]

Theorem 3 (Extreme value theorem). If \(f\) is continuous on \([a,b]\), then there is some number \(y\) in \([a,b]\) such that \(f(y)\geq f(x)\) for all \(x\in[a,b]\).

Proof of the Extreme value theorem

Proof idea.

• By Theorem 7-2, the set \( A = \{f(x) : a\leq x\leq b\}\) is bounded.
• Set \(\alpha = \sup A.\)
• If \(f(x)\neq \alpha\) for all \(x\) in \([a,b]\), then the function \[g(x) = \frac{1}{\alpha-f(x)}\] is well-defined and continuous on \([a,b]\).
• By Theorem 7-2, \(g\) is bounded above on \([a,b]\).
• Note \(N>0\) and \[g(x)=\frac{1}{\alpha-f(x)}\leq N\ \Leftrightarrow\ f(x)\leq \alpha-\frac{1}{N}\]

The Archimedean property

Theorem 8-2. The set \(\N\) is not bounded above.

Part 9:

Derivatives

(Read Chapter 9 in the textbook)

"Locally linear"

What does it mean to say \(f\) is "locally linear" at \(a\)?

Definition. The function \(f\) is differentiable at \(a\) if \[\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\text{ exists.}\]

In this case the limit is denoted by \(f'(a)\) and is call the  derivative of \(f\) at \(a\). (We also say that \(f\) is differentiable if \(f\) is differentiable at \(a\) for every \(a\) in the domain of \(f\).)

Recall the following useful fact:

If \(f(x) = g(x)\) for all \(x\neq a\), and  \(\displaystyle{\lim_{x\to a}f(x) =L}\), then \(\displaystyle{\lim_{x\to a}g(x) =L}\).

Example. For any numbers \(c\) and \(d\), consider the function         \(f(x) = cx+d\). For any number \(a\)

\[\lim_{h\to 0}\frac{f(a+h)-f(a)}{h} = \lim_{h\to 0} \frac{c(a+h)+d-ca-d}{h} = \lim_{h\to 0}\frac{ch}{h} = \lim_{h\to 0}c = c.\]

Thus, \(f'(a) = c\).

Example. Consider the function \(S(x) = x^2\) and any number \(a\). Then

\[\lim_{h\to 0}\frac{S(a+h)-S(a)}{h} = \lim_{h\to 0}\frac{(a+h)^2-a^2}{h} = \lim_{h\to 0} \frac{2ah+h^2}{h}\]

\[ = \lim_{h\to 0}(2a+h) = 2a.\]

Thus, \(S'(a) = 2a\) for any \(a\).

Theorem 9-1. If \(f\) is differentiable at \(a\), then \(f\) is continuous at \(a\).

Differentiable \(\Rightarrow\) continuous

Continuous \(\not\Rightarrow\) differentiable

Example. Consider the function \(A(x) = |x|\).

First, we claim that \(A\) is continuous:

A pathological example

Another pathological example

Example. Consider the function

\[f(x) = \begin{cases} x^2\sin\left(\frac{1}{x}\right) & x\neq 0,\\ 0 & x=0.\end{cases}\]

Some definitions

Given a function \(f\), the function \(f'\) is called the derivative of \(f\). The domain of \(f'\) is the set of points \(a\) such that \(f\) is differentiable at \(a\), and

\[f'(a) = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.\]

Given a function \(f\), the limit

\[\lim_{h\to 0^{+}}\frac{f(a+h)-f(a)}{h}\]

is the right-hand derivative of \(f\) at \(a\), and

\[\lim_{h\to 0^{-}}\frac{f(a+h)-f(a)}{h}\]

is the left-hand derivative of \(f\) at \(a\).

Part 10:

Differentiation

(Read Chapter 10 in the textbook)

Our first goal is to show that for each \(n\in\N\), the function 

\[f(x) = x^{n}\]

is differentiable on all of \(\R\), and

\[f'(x) = nx^{n-1}.\]

(Note that we interpret \(x^{0}\) as \(1\).)

Theorem 10-3. If \(f\) and \(g\) are differentiable at \(a\), then \(f+g\) is differentiable at \(a\) and

\[(f+g)'(a) = f'(a)+g'(a).\]

Theorem 10-4. If \(f\) and \(g\) are differentiable at \(a\), then \(f\cdot g\) is differentiable at \(a\) and

\[(f\cdot g)'(a) = f'(a)g(a)+f(a)g'(a).\]

Theorem 10-6. If \(f(x) = x^{n}\) for some natural number \(n\), then

\[f'(a) = na^{n-1}\quad\text{for all numbers }a.\]

Derivatives of sums and products

The Chain Rule

Theorem 10-9. If \(g\) is differentiable at \(a\), and \(f\) is differentiable at \(g(a)\), then \(f\circ g\) is differentiable at \(a\) and

Part 11:

Significance of the Derivative

(Read Chapter 11 in the textbook)

Definition. Let \(f\) be a function and \(A\) a set of numbers contained in the domain of \(f\). A point \(x\) in \(A\) is a maximum point for \(f\) on \(A\) if 

\[f(x)\geq f(y)\quad\text{for every }y\text{ in }A.\]

The number \(f(x)\) itself is called the maximum value of \(f\) on \(A\) (and we also say that \(f\) "has it's maximum value on \(A\) at \(x\).")

Extrema an critical points

Example 1

Example 2

Extrema and derivatives

Theorem 11-1. Let \(f\) be any function defined on \((a,b)\). If \(x\) is a maximum (or a minimum) point for \(f\) on \((a,b)\), and \(f\) is differentiable at \(x\), then \(f'(x) = 0\).

Theorem (Rolle's Theorem). If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), and \(f(a)=f(b)\), then there is a number \(x\) in \((a,b)\) such that \(f'(x) = 0\).

The "baby" mean value theorem

Theorem (The Mean Value Theorem). If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there is a number \(x\) in \((a,b)\) such that

\[f'(x) = \frac{f(b)-f(a)}{b-a}.\]

The Mean Value Theorem

Presentation 1:

Uniform continuity

(Read Chapter 8, Appendix in the textbook)

Definition. The function \(f\) is uniformly continuous on an interval \(A\) if for every \(\varepsilon>0\) there is some \(\delta>0\), such that, for all \(x\) and \(y\) in \(A\),

\[\text{if }|x-y|<\delta\text{, then }|f(x)-f(y)|<\varepsilon.\]

back to Part 11...

Some differential equation theorems

Corollary 11-1. If \(f\) is defined on an interval and \(f'(x)=0\) for all \(x\) in the interval, then \(f\) is constant on the interval.

Note that this shows that the only solution to the differential equation

\[\frac{dy}{dx} = 0\]

is a constant function.

Corollary 11-2. If \(f\) and \(g\) are defined on the same interval, and \(f'(x) = g'(x)\) for all \(x\) in the interval, then there is a number \(c\) such that \(f(x) = g(x)+c\) for all \(x\) in the interval.

Increasing & Decreasing functions

Corollary 11-3. If \(f'(x)>0\) for all \(x\) in some interval, then \(f\) is increasing on the interval. If \(f'(x)<0\) for all \(x\) in some interval, then \(f\) is decreasing on the interval.

Part 12:

Integrals

(Read Chapter 13 in the textbook)

Definition. Let \(a<b\). A partition of \([a,b]\) is a finite set \(P\subset [a,b]\) such that \(\{a,b\}\subset P.\)

Example. The sets \(P=\{0,\frac{1}{2},1\}\) and \(Q=\{0,\frac{1}{4},\frac{3}{4},1\}\) are partions of \([0,1]\).

Example. Given \(a<b\) and \(n\in\N\) the set

\[P = \left\{a,a+\frac{b-a}{n},a+2\frac{b-a}{n},\ldots,a+(n-1)\frac{b-a}{n},b\right\}\]

is an important example of a partition of \([a,b]\).

Example. Partitions don't have to be evenly spaced! The set

\[P=\{0,0.1,0.3,0.33,0.52,0.661,0.87,1\}\]

is a partition of \([0,1]\).

Important! When we write that \(P=\{t_{0},t_{1},\ldots,t_{n}\}\) is a partition of \([a,b]\) we will assume that the points are in the following order

\[a=t_{0}<t_{1}<\cdots<t_{n} = b.\]

Definition. Given a bounded function \(f\) on \([a,b]\) and a partition \(P = \{t_{0},t_{1},\ldots,t_{n}\}\), for each \(i\in\{1,2,3,\ldots,n\}\) we set

\[m_{i} = \inf\{f(x) : t_{i-1}\leq x\leq t_{i}\}\]

and

\[M_{i} = \sup\{f(x) : t_{i-1}\leq x\leq t_{i}\}.\]

The lower sum of \(f\) for \(P\) is the number

\[L(f,P) = \sum_{i=1}^{n}m_{i}(t_{i}-t_{i-1}).\]

The upper sum of \(f\) for \(P\) is the number

\[U(f,P) = \sum_{i=1}^{n}M_{i}(t_{i}-t_{i-1}).\]

We consider a function \(f\) defined on \([a,b]\)

Example. Consider the function \(f(x) = \frac{1}{2}x+1\) on the interval \([0,2]\). For \(n\in\mathbb{N}\) we consider the partition

\[P_{n} = \left\{0,\frac{2}{n},\frac{4}{n},\frac{6}{n},\ldots,\frac{2n-2}{n},2\right\}.\]

Note that for each \(i\in\{1,2,\ldots,n\}\) we have \[m_{i} = \inf\{f(x) : \tfrac{2i-2}{n}\leq x\leq \tfrac{2i}{n}\} = f(\tfrac{2i-2}{n}) = \frac{i-1+n}{n},\]

and \[M_{i} = \sup\{f(x) : \tfrac{2i-2}{n}\leq x\leq \tfrac{2i}{n}\} = f(\tfrac{2i}{n}) = \frac{i+n}{n}.\]

Thus, \[L(f,P_{n}) = \sum_{i=1}^{n}m_{i}\left(\frac{2i}{n} - \frac{2i-2}{n}\right) = \sum_{i=1}^{n}\frac{i-1+n}{n}\left(\frac{2}{n}\right).\]

\[ = \frac{2}{n^{2}}\sum_{i=1}^{n}(i-1+n) = \frac{2}{n^{2}}\left(\sum_{i=1}^{n}i - \sum_{i=1}^{n}1 + \sum_{i=1}^{n}n\right)\]

\[ =\frac{2}{n^{2}}\left(\frac{n(n+1)}{2} - n + n^2\right) = \frac{2}{n^{2}}\left(\frac{3}{2}n^{2}-\frac{1}{2}n\right) = 3-\frac{1}{n}\]

Example continued. Similarly, we can calculate

\[U(f,P_{n}) = 3+\frac{1}{n}.\]

Lemma. Suppose \(f\) is a bounded function on \([a,b]\). If \(P\) and \(Q\) are partitions of \([a,b]\) and \(P\subset Q\), then

\[L(f,P)\leq L(f,Q)\quad\text{and}\quad U(f,P)\geq U(f,Q).\]

Our first goal is to show that for any two partitions \(P_{1}\) and \(P_{2}\) of \([a,b]\) we always have

\[L(f,P_{1})\leq U(f,P_{2}).\]

First, we need a lemma:

Proof continued. Note that \[L(f,P) = \sum_{i=1}^{n}m_{i}(t_{i}-t_{i-1})\] and \[L(f,Q) = \sum_{i=1}^{k}m_{i}(t_{i}-t_{i-1}) +m'(u-t_{k})+m''(t_{k+1}-u) + \sum_{i=k+2}^{n}m_{i}(t_{i}-t_{i-1}).\] 

Thus, to show that \(L(f,P)\leq L(f,Q)\) we must show that

\[m_{k+1}(t_{k+1}-t_{k})\leq m'(u-t_{k})+m''(t_{k+1}-u).\]

Note that \(\{f(x) : t_{k}\leq x\leq u\}\) and \(\{f(x) : u\leq x\leq t_{k+1}\}\) are both subsets of \(\{f(x) : t_{k}\leq x\leq t_{k+1}\}\). This implies that the infimums of the first two sets, that is, \(m'\) and \(m''\) can be no smaller than the infimum \(m_{k+1}\) of the third set, and thus \[m_{k+1}\leq m'\quad\text{and}\quad m_{k+1}\leq m''.\]

Therefore, \[m_{k+1}(t_{k+1}-t_{k}) = m_{k+1}(u-t_{k}) + m_{k+1}(t_{k+1}-u)\]

\[ \leq m'(u-t_{k})+m''(t_{k+1}-u).\]

Proof continued. Now, since \(P\subset Q\) and both are finite sets, there is a sequence of partitions

\[P = P_{0}\subset P_{1}\subset P_{2}\subset\cdots\subset P_{m} = Q\]

where each  \(P_{m+1}\) contains exactly one more point than \(P_{m}\), and thus

\[L(f,P) = L(f,P_{0})\leq L(f,P_{1})\leq L(f,P_{2})\leq \cdots\leq L(f,P_{m}) = L(f,Q).\]

The proof that \(U(f,P)\geq U(f,Q)\) is similar and you should fill in the details. \(\Box\)

Theorem 1. If \(f\) is a bounded function on \([a,b]\) and \(P_{1}\) and \(P_{2}\) are two partitions of \([a,b]\), then

\[L(f,P_{1})\leq U(f,P_{2}).\]

Proof. Let \(Q = P_{1}\cup P_{2}\), then \(Q\) is a partition of \([a,b]\) such that both \(P_{1}\subset Q\) and \(P_{2}\subset Q\). Clearly we have \(L(f,Q)\leq U(f,Q).\) Thus, by the lemma

\[L(f,P_{1})\leq L(f,Q)\leq U(f,Q)\leq U(f,P_{2}).\quad\Box\]

Given a function \(f\) which is bounded on \([a,b]\) we consider the set of all lower sums:

\[\{L(f,P)\}:=\{L(f,P) : P\text{ is a partition of }[a,b]\}\]

and the set of all upper sums:

\[\{U(f,P)\}:=\{U(f,P): P\text{ is a partition of }[a,b]\}\]

Consider the partition \(Q = \{a,b\}\), then for any partition \(P\) of \([a,b]\) we have

\[L(f,P)\leq U(f,Q) = (b-a)\sup\{f(x) : a\leq x\leq b\}\]

and

\[U(f,P)\geq L(f,Q)  = (b-a)\inf\{f(x) : a\leq x\leq b\}.\]

We conclude that

\[\sup\{L(f,P)\} \leq \inf\{U(f,P)\}.\]

In particular, the supremum and infimum above exist.

Example. Let \(g:\mathbb{R}\to\mathbb{R}\) be given by

\[g(x) = \begin{cases} 1 & x\in\mathbb{Q},\\ 0 & x\in\mathbb{R}\setminus\mathbb{Q}.\end{cases}\]

Let \(a<b\) and let\(P = \{t_{0},t_{1},\ldots,t_{n}\}\) be a partition of \([a,b]\). For each \(i\in\{1,2,\ldots,n\}\) there is a rational number \(q\in\mathbb{Q}\) such that \(t_{i-1}<q<t_{i}\) and there is an irrational number \(z\in\mathbb{R}\setminus\mathbb{Q}\) such that \(t_{i-1}<z<t_{i}\). From this we see that \[m_{i} = \inf\{g(x) : t_{i-1}\leq x\leq t_{i}\} = 0\] and \[M_{i} = \sup\{g(x) : t_{i-1}\leq x\leq t_{i}\} = 1.\]

Thus \[L(g,P) = \sum_{i=1}^{n}m_{i}(t_{i}-t_{i-1}) = 0\] and \[U(g,P) = \sum_{i=1}^{n} M_{i}(t_{i}-t_{i-1}) = \sum_{i=1}^{n}(t_{i}-t_{i-1}) = t_{n}-t_{0} = b-a.\]

Example continued. From this, we see that

\[\{L(g,P)\} = \{0\}\] and \[\{U(g,P)\} = \{b-a\}.\]

Thus, we see that 

\[\sup\{L(g,P)\} = 0< b-a = \inf\{U(g,P)\}.\]

Definition. A function \(f\) which is bounded on \([a,b]\) is integrable on \([a,b]\) if

\[\sup\{L(f,P) : P\text{ is a partition of }[a,b]\}\]

\[\qquad\qquad = \inf\{U(f,P) : P\text{ is a partition of }[a,b]\}.\]

In this case, the common number is called the integral of \(f\) on \([a,b]\) and is denoted with the symbol

\[\int_{a}^{b}f.\]

Example continued. The function \(g\) is not integrable on \([a,b]\) for any \(a<b\).

Example (constant function). Let \(c\in\mathbb{R}\), and define the function \(h:\mathbb{R}\to\mathbb{R}\) were \(h(x) = c\) for all \(x\in\mathbb{R}\).

Let \(a<b\), and let \(P = \{t_{0},t_{1},\ldots,t_{n}\}\) be a partition of \([a,b]\), then \[m_{i} = c = M_{i}\], and hence

\[L(h,P) = U(h,P) = \sum_{i=1}^{n}c(t_{i}-t_{i-1}) = c(b-a).\]

Thus,

\[\int_{a}^{b}h = c(b-a).\]

Example (a linear function). Recall the function \(f(x) = \frac{1}{2}x+1\) on the interval \([0,2]\). For \(n\in\mathbb{N}\) we consider the partition

\[P_{n} = \left\{0,\frac{2}{n},\frac{4}{n},\frac{6}{n},\ldots,\frac{2n-2}{n},2\right\}.\]

Note that for each \(i\in\{1,2,\ldots,n\}\) we have \[m_{i} = \inf\{f(x) : \tfrac{2i-2}{n}\leq x\leq \tfrac{2i}{n}\} = f(\tfrac{2i-2}{n}) = \frac{i-1+n}{n},\]

and \[M_{i} = \sup\{f(x) : \tfrac{2i-2}{n}\leq x\leq \tfrac{2i}{n}\} = f(\tfrac{2i}{n}) = \frac{i+n}{n}.\]

We saw that

\[L(f,P_{n}) = 3-\frac{1}{n}\quad\text{and}\quad U(f,P_{n}) = 3+\frac{1}{n}.\]

Since \[\left\{2,\frac{5}{2},\frac{8}{3},\ldots, 3-\frac{1}{k},\ldots\right\}\subset \{L(f,P) : P\text{ is a partition of }[a,b]\}\]

We see that \[\sup\{L(f,P) : P\text{ is a partition of }[a,b]\}\geq 3\]

Example (a linear function). Similarly, 

Since \[\left\{4,\frac{7}{2},\frac{10}{3},\ldots, 3+\frac{1}{k},\ldots\right\}\subset \{U(f,P) : P\text{ is a partition of }[a,b]\}\]

We see that \[\inf\{U(f,P) : P\text{ is a partition of }[a,b]\}\leq 3.\]

Putting these together we have \[3\leq \sup\{L(f,P)\}\leq \inf\{U(f,P)\}\leq 3\]

and thus \[\sup\{L(f,P)\}=\inf\{U(f,P)\} = 3.\]

This shows \[\int_{0}^{2}f = 3.\]

Note: We will also write this integral as:

\[\int_{0}^{2}\frac{1}{2}x+1\,dx\qquad\text{or}\qquad \int_{0}^{2}\frac{1}{2}w+1\,dw\]

or ANY other variable in place of \(w\).

Theorem 13-3. Suppose \(f\) is bounded on \([a,b]\). The function \(f\) is integrable on \([a,b]\) if and only if for every \(\varepsilon>0\) there exists a partition \(P\) of \([a,b]\) such that

\[U(f,P)-L(f,P)<\varepsilon.\]

Corollary. Suppose \(f\) is bounded on \([a,b]\). If there is a sequence \(\{P_{n}\}\) of partitions of \([a,b]\) such that

\[\lim_{n\to\infty}(U(f,P_{n})-L(f,P_{n})) = 0,\]

then \(f\) is integrable and

\[\int_{a}^{b} f = \lim_{n\to\infty}U(f,P_{n}) = \lim_{n\to\infty}L(f,P_{n}).\]

Proof. In the text. \(\Box\)

Proof. From the definition of the limit we see that for every \(\varepsilon>0\) there exists \(n\in\N\) such that \(U(f,P_{n})-L(f,P_{n})<\varepsilon.\) Thus, by Theorem 3, the function \(f\) is integrable on \([a,b]\). Let

\[I = \int_{a}^{b}f.\]

Then, for each \(n\in\N\) we have \[0\leq I-L(f,P_{n})\leq U(f,P_{n})-L(f,P_{n}).\]

For arbitrary \(\varepsilon>0\) there exists \(N\in\N\) such that \(|U(f,P_{n})-L(f,P_{n})|<\varepsilon\) for all \(n\geq N\), and hence for \(n\geq N\) we have

\[|I-L(f,P_{n})| = I-L(f,P_{n})\leq U(f,P_{n})-L(f,P_{n})<\varepsilon.\]

This shows that the sequence \(\{L(f,P_{n})\}\) converges to \(I\). The proof that \(\{U(f,P_{n})\}\) converges to \(I\) is left to you! \(\Box\)

Example. Let \(a<c<b\) and define the function \(\delta_{c}:\mathbb{R}\to\mathbb{R}\) by

\[\delta_{c}(x) = \begin{cases} 1 & x=c\\ 0 & \text{otherwise}.\end{cases}\]

For each \(n\in\mathbb{N}\) define the partition \(P_{n}\) of \([a,b]\) by

\[P_{n} = \left\{a,c-\frac{c-a}{2n},c+\frac{b-c}{2n},b\right\}.\]

Our next goal is to prove the following theorems:

Theorem 5. If \(f\) and \(g\) are both integrable on \([a,b]\), then \(f+g\) is integrable on \([a,b]\) and

\[\int_{a}^{b}(f+g) = \int_{a}^{b}f + \int_{a}^{b} g.\]

Theorem 6. If \(f\) is integrable on \([a,b]\) and \(c\in\mathbb{R}\), then \(cf\) is integrable on \([a,b]\) and

\[\int_{a}^{b}cf = c\int_{a}^{b}f.\]

Together, they will allow us to greatly increase the set of functions that we can integrate.

Presentation 2:

Inverse functions

(Read Chapter 12 in the textbook)

Theorem 12-1. \(f^{-1}\) is a function if and only if \(f\) is one-one.

Some preliminaries

Definitions: A function if one-one if \(f(a)=f(b)\) implies \(a=b\).

Given a function \(f\), the set \(f^{-1}\) is the set of all ordered pairs \((a,b)\) such that \((b,a)\) is in \(f\).

Note that if \(f(0)=f(1)=2\), then the symbol \(f^{-1}(2)\) is not well-defined. This observation leads to the following theorem:

Some preliminaries

Some preliminaries

Proof. Case 2: \(f(c)<f(a)<f(b)\)

Some preliminaries

Proof. We've seen:

Case 1: \(f(a)<f(c)<f(b)\) 

Case 2: \(f(c)<f(a)<f(b)\)

The rest of the cases are similar:

Case 3: \(f(b)<f(c)<f(a)\)

Case 4: \(f(b)<f(a)<f(c)\).

Some preliminaries

Proof. Fix numbers \(a<b\)  in the interval.

Case 1: Suppose \(f(a)<f(b)\). Let let \(c\) and \(d\) be any two numbers in the interval with \(c<d\). By Lemma 2, no matter what order the numbers \(\{a,b,c,d\}\) are in, we see that \(f(c)<f(d)\). Since \(c\) and \(d\) were arbitrary, we conclude \(f\) is increasing.

Case 2: Suppose \(f(a)>f(b)\).  Let let \(c\) and \(d\) be any two numbers in the interval with \(c<d\). By Lemma 2, no matter what order the numbers \(\{a,b,c,d\}\) are in, we see that \(f(c)>f(d)\). Since \(c\) and \(d\) were arbitrary, we conclude \(f\) is decreasing. \(\Box\).

Ray's presentation:

Lemma. If \(f\) is continuous and one-one on an interval \(A\), then the set

\[B = \{f(x) : x\in A\}\]

is an interval.

Theorem 12-3. If \(f\) is continuous an one-one on an interval \(A\), then \(f^{-1}\) is continuous.

Lemma. If \(f\) and \(g\) are bounded functions on \([a,b]\), and \(P\) is a partition of \([a,b]\), then

\[L(f,P)+L(g,P)\leq L(f+g,P)\leq U(f+g,P)\leq U(f,P)+U(g,P),\]

and thus

\[\sup\{L(f,P)\} + \sup\{L(g,P)\}\leq \sup\{L(f+g,P)\}\]

and

\[\inf\{U(f+g,P)\} \leq \inf\{U(f,P)\}+ \inf\{U(g,P)\}.\]

Proof continued. This implies that

\[m_{i}+m_{i}'\leq m_{i}''.\]

A similar proof (you should fill in the details) shows 

\[M_{i}''\leq M_{i}+M_{i}'.\]

Using this, we have

\[= \sum_{i=1}^{n}(m_{i}+m_{i}')(t_{i}-t_{i-1}) \leq \sum_{i=1}^{n}m_{i}''(t_{i}-t_{i-1}) = L(f+g,P)\]

\[L(f,P)+L(g,P) = \sum_{i=1}^{n}m_{i}(t_{i}-t_{i-1})+\sum_{i=1}^{n}m_{i}'(t_{i}-t_{i-1}) \]

and

\[U(f+g,P) =\sum_{i=1}^{n}M_{i}''(t_{i}-t_{i-1}) \leq \sum_{i=1}^{n}(M_{i}+M_{i}')(t_{i}-t_{i-1})\]

\[=\sum_{i=1}^{n}M_{i}(t_{i}-t_{i-1}) + \sum_{i=1}^{n}M_{i}'(t_{i}-t_{i-1}) = U(f,P)+U(g,P)\]

Proof continued. We have shown that for any partition \(P\) we have

\[L(f,P)+L(g,P)\leq L(f+g,P).\]

Let \(\varepsilon>0\) be arbitrary. Let \(P_{1}\) and \(P_{2}\) be partitions of \([a,b]\) such that

\[L(f,P_{1})>\sup\{L(f,P)\}-\frac{\varepsilon}{2}\quad\text{and}\quad L(g,P_{2})>\sup\{L(f,P)\}-\frac{\varepsilon}{2}.\]

Set \(Q=P_{1}\cup P_{2}\), then we have

\[\leq L(f,Q)+L(g,Q)\leq L(f+g,Q).\]

\[\sup\{L(f,P)\}+\sup\{L(g,P)\}-\varepsilon < L(f,P_{1}) + L(g,P_{2})\]

\[\leq \sup\{L(f+g,P)\}\]

Thus, for any \(\varepsilon>0\) we have

\[\sup\{L(f,P)\}+\sup\{L(g,P)\}<\sup\{L(f+g),P\}+\varepsilon.\]

The only possibility is that

\[\sup\{L(f,P)\}+\sup\{L(g,P)\}\leq \sup\{L(f+g),P\}.\]

The proof of the other inequality is left to you! \(\Box\)

Theorem 13-5. If \(f\) and \(g\) are both integrable on \([a,b]\), then \(f+g\) is integrable on \([a,b]\) and

\[\int_{a}^{b}(f+g) = \int_{a}^{b}f + \int_{a}^{b} g.\]

Proof. Since \(f\) and \(g\) are both integrable on \([a,b]\) we have

\[\sup\{L(f,P)\} = \inf\{U(f,P)\}\quad\text{and}\quad \sup\{L(g,P)\} = \inf\{U(g,P)\}.\]

By the previous lemma

Since the left and right ends of this inequality are equal, we conclude that all of the inequalities are equalities. That the middle inequality is an equality shows that \(f+g\) is integrable on \([a,b]\). The other equalities give the desired formula for the integral of \(f+g\). \(\Box\)

\[\sup\{L(f,P)\} + \sup\{L(g,P)\}\leq \sup\{L(f+g,P)\}\]

\[\leq \inf\{U(f+g,P)\} \leq \inf\{U(f,P)\} + \inf\{U(g,P)\}\]

Theorem 13-6. If \(f\) is integrable on \([a,b]\) and \(c\in\mathbb{R}\), then \(cf\) is integrable on \([a,b]\) and

\[\int_{a}^{b}cf = c\int_{a}^{b}f.\]

Proof. This proof is similar to the proof of Theorem 5. You should fill in the details. One hint is to consider the cases of \(c\geq 0\) and \(c\leq 0\) separately. \(\Box\)

Corollary. If \(f_{1},f_{2},\ldots,f_{n}\) are all integrable on \([a,b]\) and \(a_{1},a_{2},\ldots,a_{n}\) is a list of real numbers, then

\[f = a_{1}f_{1} + a_{2}f_{2} + \cdots + a_{n}f_{n}\]

is integrable on \([a,b]\), and

\[\int_{a}^{b}f =a_{1}\int_{a}^{b}f_{1} + a_{2}\int_{a}^{b}f_{2} + \cdots + a_{n}\int_{a}^{b}f_{n}.\]

Proof. Use induction with Theorems 5 and 6. \(\Box\)

Theorem 13-3. If \(f\) is continuous on \([a,b]\) then \(f\) is integrable on \([a,b].\)

Many theorems require us to know that a function is integrable. It's enough that the function is continuous...

Continuous \(\Rightarrow\) integrable

Recall Theorem 1 Appendix 8:

Theorem 1. If \(f\) is continuous on \([a,b]\) then \(f\) is uniformly continuous on \([a,b]\).

Example. In the text and in homework we have/will see that constant functions, \(x\), \(x^{2}\) and \(x^{3}\) are all integrable on \([0,b]\). Using theorems 5 and 6 we see that any polynomial of degree \(\leq 3\) is integrable on \([0,b]\).

Example. Consider the functions \(g:[0,1]\to\mathbb{R}\) given by

\[g(x) = \begin{cases} 1 & \text{if }\frac{1}{x}\in\mathbb{N},\\ 0 & \text{otherwise}.\end{cases}\]

Theorem 13-4. Let \(a<c<b\). If \(f\) is integrable on \([a,b]\), then \(f\) is integrable on \([a,c]\) and on \([c,b]\). Conversely, if \(f\) is integrable on \([a,c]\) and on \([c,b]\), then \(f\) is integrable on \([a,b]\). Finally, if \(f\) is integrable on \([a,b]\) then

\[\int_{a}^{b}f = \int_{a}^{c}f + \int_{c}^{b}f.\]

On the other hand, if \(a\leq y<x\leq b\) then

\[F(y)-F(x) = \int_{a}^{y}f-\int_{a}^{x}f = \int_{a}^{y}f-\left(\int_{a}^{y}f +\int_{y}^{x}f\right)= -\int_{y}^{x}f.\]

Hence, for \(y<x\) we define 

\[\int_{x}^{y}f = -\int_{y}^{x}f.\]

Thus, no matter which order \(x\) and \(y\) are in, we have

\[F(y)-F(x) = \int_{x}^{y}f.\]

Theorem 13-7. If \(f\) is integrable on \([a,b]\) and

\[m\leq f(x)\leq M\quad\text{for all }x\in[a,b],\] then \[m(b-a)\leq\int_{a}^{b}f\leq M(b-a).\]

Theorem 13-8. If \(f\) is integrable on \([a,b]\), then \(F:[a,b]\to\mathbb{R}\) given by \[F(x) = \int_{a}^{x}f\] is continuous on \([a,b]\).

Example.  \[f(x) = \begin{cases} 0 & x<c,\\ 1 & x\geq c.\end{cases}\]

 \[F(x) = \int_{a}^{x}f = \begin{cases} 0 & x\leq c,\\ x-c & x\geq c.\end{cases}\]

Note that \(F\) is not differentiable at \(c\). The next theorem shows us that if \(f\) were continuous, then \(F\) would be differentiable.

Part 13:

The Fundamental Theorem of Calculus

(Read Chapter 14 in the textbook)

Theorem 14-1 (the fundamental theorem of calculus). Let \(f\) be integrable on \([a,b]\) and define \(F:[a,b]\to\mathbb{R}\) by

\[F(x) = \int_{a}^{x}f.\]

If \(f\) is continuous at \(c\) in \((a,b)\), then \(F\) is differentiable at \(c\), and

\[F'(c) = f(c).\]

Moreover, if \(f\) is continuous at \(a\) or \(b\), then

\[\lim_{h\to 0^{+}}\frac{F(a+h)-F(a)}{h} = f(a)\quad\text{or}\quad \lim_{h\to 0^{-}}\frac{F(b+h)-F(b)}{h} = f(b),\]

respectively.

Note: for \(F\) to be differentiable it is sufficient that \(f\) is continuous. If \(f=\delta_{c}\) for some \(c\in[a,b]\), then \(f\) is not continuous, but we see that \(F\) is the zero function, which is differentiable. Also note that \(F'(c)=0\neq f(c)\).

Show that for \(0<a<b\) and \(p\in\N\)

\[\int_{a}^{b}x^p\,dx  = \frac{1}{p+1}(b^{p+1}-a^{p+1}).\]

See Problem 4 in Chapter 13 for an outline.

3rd presentation:

Show that for \(c>1\), we have

\[\lim_{n\to\infty}c^{\frac{1}{n}} = 1,\]

that is, for every \(\varepsilon>0\), there exists a natural number \(M\) such that if \(n\geq M\) and \(n\in \N\), then \(|c^{\frac{1}{n}}-1|<\varepsilon.\) (See Problem 41(a) in Chapter 5.)

4th presentation:

Corollary 14-2. Suppose \(f\) is continuous on \([a,b]\) and for some function \(g\) we have \(f(x) = g'(x)\) for all \(x\in(a,b)\), then

\[\int_{a}^{b}f = g(b) - g(a).\]

Proof. Define

\[F(x) = \int_{a}^{x}f.\]

By the fundamental theorem of calculus we have \(F'(x) = f(x) = g'(x)\) for all \(x\in(a,b)\). By the Corollary 11-2, there exists \(C\in\mathbb{R}\) such that \(F(x) = g(x)+C\) for all \(x\in[a,b]\), and thus

\[\int_{a}^{b}f = F(b)-F(a) = (g(b)+C)-(g(a)+C) = g(b) - g(a).\ \Box\]

Example. For \(a\leq b\) we wish to evaluate \[\int_{a}^{b}x^{4}\,dx\]

Define the function \[F(x) = \int_{a}^{x}t^{4}\,dt,\]

and \(g(x) = x^{5}/5\). Note that \(F'(x) = x^{4} = g'(x)\) for all \(x\in (a,b)\), and hence by Corollary 14-2 we have

\[\int_{a}^{b}x^{4}\,dx = g(b)-g(a) = \frac{b^{5}-a^{5}}{5}.\]

Theorem. Let \(a_{0},a_{1},\ldots,a_{n}\) be real numbers. If

\[f(x) = a_{0}+a_{1}x+\cdots+a_{n}x^{n},\]

and \(a\leq b\), then

\[\int_{a}^{b} f = a_{0}(b-a) + a_{1}\frac{b^{2}-a^{2}}{2} + a_{2}\frac{b^{3}-a^{3}}{3} + \cdots + a_{n}\frac{b^{n+1}-a^{n+1}}{n+1}.\]

We can generalize the previous example in order to integrate any polynomial on a closed interval.

Part 14:

Logs and exponents

(Read Chapter 18 in Book of Proof)

Example continued. We can define the function \(F:(0,\infty)\to\mathbb{R}\) by

\[F(x) = \int_{1}^{x}\frac{1}{t}\,dt.\]

We know from the fundamental theorem of calculus that \(F\) is differentiable on \((0,\infty)\), but is it a function we've heard of before? YES!

The inverse of \(\log\)

Observe that \(\log:(0,\infty)\to\mathbb{R}\) is continuous.

Moreover, for any \(n\in\mathbb{N}\) we have

\[\log(2^{n}) = n\log(2)\]

and

\[\log\left(\frac{1}{2^{n}}\right) = -n\log(2).\]

Since \[\log(2) = \int_{1}^{2}\frac{1}{t}\,dt>0\] we see that \(\log\) is not bounded above or below.

Definition. The exponential function, \(\operatorname{exp} = \log^{-1}:\R\to (0,\infty)\)

The exponential function

Theorem 18-2. For each \(x\in\mathbb{R}\),

\[\exp'(x) = \exp(x).\]

Theorem 18-3. For each \(x,y\in\mathbb{R}\),

\[\exp(x+y) = \exp(x)\cdot\exp(y).\]

Theorem 12-5. Let \(f\) be continuous one-one function defined on an interval, and suppose that \(f\) is differentiable at \(f^{-1}(b)\), and \(f'(f^{-1}(b)) \neq 0\). Then \(f^{-1}\) is differentiable at \(b\) and \[(f^{-1})'(b) = \frac{1}{f'(f^{-1}(b))}.\]

Real exponents

Definition. 

\[e: = \exp(1).\]

Note that for \(n\in\N\) we have

\[e^n = \exp(1)\cdot\exp(1) \cdots \exp(1)= \exp(1+1+\cdots+1) = \exp(n)\]

Note that \(e\) is the (unique) number such that

\[\int_{1}^{e}\frac{1}{x}\,dx = 1.\]

Let's use our definition to estimate \(e\).

Theorem 18-4. If \(a>0\), then

\[(a^b)^c = a^{bc}\quad\text{for all }b, c,\]

\[a^{1} = a,\]

and 

\[a^{x+y} = a^x\cdot a^{y}\quad\text{for all }x,y.\]

Part 15:

Lebesgue Measure

Generalizing upper and lower sums

Consider the definition of upper and lower sums:

1. Break the interval \([a,b]\) into subintervals: \[[a,t_{1}], [t_{1},t_{2}],[t_{2},t_{3}],\ldots,[t_{n-1},b].\]
2. Compute an upper and lower bound for the "area" under the function \(f\) on the subinterval: \[\text{lower bound}: \inf\{f(x) : x\in[t_{i-1},t_{i}]\} (t_{i}-t_{i-1})\]  \[\text{upper bound}: \sup\{f(x) : x\in[t_{i-1},t_{i}]\} (t_{i}-t_{i-1})\]
3. Add up the lower bounds and upper bounds.

The "length" of set

What "should" be the length of these subsets of \(\mathbb{R}\):

set              length

\([a,b]\) w/ \(a\leq b\)

Sets with measure zero

Definition. Let \(A\) be a set of real numbers. We say that \(A\) has measure zero if for each \(\varepsilon>0\) there exists a sequence of open intervals \((a_{1},b_{1}),(a_{2},b_{2}),(a_{3},b_{3}),\ldots\) such that
\[A\subset \bigcup_{n=1}^{\infty}(a_{n},b_{n})\quad\text{and}\quad \sum_{n=1}^{\infty}(b_{n}-a_{n})<\varepsilon.\]

Example 2. Consider the set \(\N\). For arbitrary \(\varepsilon>0\) and \(n\in\N\) we define the open interval

\[U_{n} = \left(n-\frac{\varepsilon}{2^{n+2}},n+\frac{\varepsilon}{2^{n+2}}\right).\]

Note that

\[\N\subset \bigcup_{n=1}^{\infty} U_{n},\]

and the "length" of \(U_{n}\) is \(\varepsilon 2^{-n-1}\), thus the sum of the lengths of the \(U_{n}\)'s is \(\varepsilon/2<\varepsilon.\)

More generally, if \(A\subset\mathbb{R}\) is a countable set, that is, there exists a surjective (onto) function \(f:\mathbb{N}\to A\), then \(A\) has measure zero.

Countable sets have measure zero

Measure of the rationals

Consider the set

\[A = \left\{0,1,\frac{0}{2},\frac{1}{2},\frac{2}{2},\frac{0}{3},\frac{1}{3},\frac{2}{3},\frac{3}{3},\frac{0}{4},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{4}{4},\ldots\right\}.\]

This set is countable, indeed, let \(f(n)\) be the \(n\) element in this list, then \(f:\mathbb{N}\to A\) is a surjection.

Note that \(f(1) = 0 = f(3)\), so this function is not one-one.

Which sets can be measured

Definition. Let \(\Sigma\) be a collection of subsets of \(\mathbb{R}\). We say that \(\Sigma\) is a   \(\sigma\)-algebra if it satisfies the following three properties:

1. \(\displaystyle{\varnothing\in\Sigma}\)
2. if \(A\in\Sigma\), then \(A^{c}:=\mathbb{R}\setminus A\in\Sigma\)
3. if \(A_{1},A_{2},\ldots,A_{n},\ldots\) is a sequence of sets such that \(A_{n}\in\Sigma\) for each \(n\in\N\), then \(\displaystyle{\bigcup_{n=1}^{\infty}A_{n}\in\Sigma}\).

The set \(\mathcal{B}\) is the smallest \(\sigma\)-algebra such that every open interval is in \(\mathcal{B}\). If \(A\in\mathcal{B}\), then \(A\) is a called a Borel set.

Consider the set \(\mathcal{X}\) of all \(\sigma\)-algebras that contain all of the open intervals \((a,b)\) such that \(a<b\).

Note that \(\mathcal{P}(\mathbb{R})\in\mathcal{X}\), so \(\mathcal{X}\) isn't empty.

Then the Borel sets is the intersection

\[\mathcal{B} = \bigcap_{\Sigma\in\mathcal{X}}\Sigma.\]

Smallest how?

Some Borel sets

Example 1. A single point set \(\{a\}\) is a Borel set since

\[\{a\} = \big((-\infty,a)\cup(a,\infty)\big)^{c}.\]

Example 2. The sets \(\N,\Z,\) and \(\mathbb{Q}\) are Borel sets. Indeed, all countable sets are Borel sets.

Theorem. If \(A_{1},A_{2},\ldots\) are Borel sets, then

\[\bigcap_{n=1}^{\infty}A_{n}\]

is a Borel set.

Proof. Note that

\[\left(\bigcap_{n=1}^{\infty}A_{n}\right)^{c} = \bigcup_{n=1}^{\infty}A_{n}^{c}.\ \Box\]

What is Lebesgue measure?

Definition. Given a \(\sigma\)-algebra \(\Sigma\), a measure on \(\Sigma\) is a function \(\mu\) which takes elements of \(\Sigma\) as inputs, outputs nonnegative real numbers, and has the following properties:

1. \(\mu(\varnothing) = 0\)
2. For any countable collection of sets \(A_{1},A_{2},A_{3},\ldots\) in \(\Sigma\) such that \(A_{n}\cap A_{m}=\varnothing\) for all \(m\neq n\), it follows that \[\mu\left(\bigcup_{n=1}^{\infty}A_{n}\right) = \sum_{n=1}^{\infty}\mu(A_{n}).\]

Lebesgue measure denoted \(m\), is the unique measure on Borel sets such that \(m((a,b)) = (b-a)\) for all \(a<b\).

Some fact about Lebesgue measure

The Lebesgue Integral

Suppose \(f\) is a function and \(A\) is a Borel set which is a subset of the domain of \(f\).

Given any finite collection of Borel sets \(A_{1},A_{2},\ldots,A_{n}\subset A\) which are pairwise disjoint and their union is \(A\), we define the upper and lower sums

\[\sum_{i=1}^{n}\big(\inf\{f(x) : x\in A_{i}\}\cdot m(A_{i})\big)\ \  \text{and}\ \ \sum_{i=1}^{n}\big(\sup\{f(x) : x\in A_{i}\}\cdot m(A_{i})\big),\] respectively.

Example. Consider the function

\[D(x) = \begin{cases} 1 & x\in\mathbb{Q}\\ 0 & x\in\mathbb{Q}^{c}.\end{cases}\]

For any Borel set \(A\) (\(A=[0,1]\) for example), we consider the sets \(A_{1} = A\cap\mathbb{Q}\) and \(A_{2} = A\cap (\mathbb{Q}^{c})\).

Since \(m(A_{1})\leq m(\mathbb{Q}) = 0\), we see that both the upper and lower sums are \(0\), thus

\[\int_{A}D = 0.\]

The Dirichlet function

Example. Consider the functions \(g:[0,1]\to\mathbb{R}\) given by

\[g(x) = \begin{cases} 1 & \text{if }\frac{1}{x}\in\mathbb{N},\\ 0 & \text{otherwise}.\end{cases}\]

In Homework 9 we showed for each integer \(n\geq 0\) and each \(x\in(0,1]\) we have
\[0< e^{x} - \left(1+\frac{x}{1!}+\frac{x^{2}}{2!}+\cdots + \frac{x^{n}}{n!}\right)< 3\frac{x^{n+1}}{(n+1)!}.\]

Thus, letting \(x=1\), for any integer \(n\geq 0\) we have

\[0< e - \left(1+\frac{1}{1!}+\frac{1}{2!}+\cdots + \frac{1}{n!}\right)< 3\frac{1}{(n+1)!}.\]

5th presentation:

Theorem 20-5. \(e\) is irrational.