Homework
No homework due on Oct 15. Recall that the Midterm is on Oct 17, in class from 9.40am  10.55am.
Homework 9 (Nov 5, 2024)
 We showed in class that if \(f:[a,b] \to \R\) is absolutely continuous, then it can be written as a difference of two nondecreasing functions. Use the same idea to prove Jordan’s theorem: \(f\) is of bounded variation iff it can be written as the difference of two nondecreasing functions.
 Let \([a,b]\) be a closed, bounded interval and the function \(f\) be of bounded variation on \([a,b]\). Show that there is an absolutely continuous function \(g\) on \([a,b]\) and a function \(h\) on \([a,b]\) that is BV and has \(h' = 0\) a.e. on \([a,b]\) for which \(f = g + h\) on \([a,b]\). Then show that this decomposition is unique except for the addition of constants.

Define \(f\) and \(g\) on \([1,1]\) by \(f(x) = x^{1/3}\) for \(1\leq x \leq 1\) and
\(g(x) = \begin{cases} x^2 \cos \left( \pi/2x \right) & x \in [1,1] \setminus \{0\} \\ 0 & x = 0 \end{cases}\)
 Show that both \(f\) and \(g\) are absolutely continuos on \([1,1]\).

For the partition \(P_n = \{1,0,1/2n,1/[2n1],\ldots,1/3,1/2,1\}\) of \([1,1]\) estimate
\(V(f,P_n) = \sum_{i} f(t_i)  f(t_{i1})\)
 Show that $f \circ g$ fails to be of bounded variation, and hence also fails to be absolutely continuous.

\(f\) is said to be Lipschitz continuous if there is a constant \(c\) such that \(\vert f(x)  f(y)\vert \leq c\vert xy\vert\) for all \(x,y \in [a,b]\). Show that \(f\) is differentiable a.e., and
\(f(x)  f(a) = \int_a^x f'\)

A function \(f\) of bounded variation is said to be singular if its derivative vanishes a.e. Suppose \(f\) is increasing on \([a,b]\) and singular. Show using the Vitali covering lemma that \(f\) has the following property: given \(\epsilon,\delta > 0\), there is a collection of intervals \(I_k = (a_k,b_k)\), \(k=1,\ldots,n\) such that
\(\sum_{k=1}^n b_k  a_k < \delta \AND \sum_{k=1}^n f(b_k)  f(a_k) > f(b)  f(a)  \epsilon\)
Homework 8 (Oct 29, 2024)
Problem 1

Let $\nu$ be a signed measure on $(X,\mathcal{F})$. Define integration over $X$ wrt $\nu$ by defining
\(\int_X f d \nu = \int_X f d \nu^+  \int_X f d \nu^{}\)
provided $f$ is integrable over $X$ with respect to both $\nu^+$ and $\nu^$. Show that if $f \leq M$ on $X$, then
\(\int_X f d \nu \leq M \nu(X).\)
Morever, if $\nu(X) < \infty$, show that there is a measurable function $f$ with $f \leq 1$ on $X$ for which
\(\int_X f d \nu = \nu(X)\)
Solution: For the first statement, we have
\(\left\int f\:d\nu\right\leq\left\int f\:d\nu^+\right+\left\int f\:d\nu^\right\leq M\int d\nu^++M\int d\nu^=M\nu(X).\)
For the second statement, take
\(f=\chi_{\text{supp }\nu^+}\chi_{\text{supp }\nu^}.\)

Let $\mu$, $\nu$ and $\lambda$ be $\sigma$finite measures on $(X,\mathcal{F})$.
 If $\nu « \mu$ and $f$ is a nonnegative function on $X$ that is measurable wrt $\mathcal{F}$, show that \(\int_X f d\nu = \int_X f \frac{d\nu}{d\mu} d\mu\)
 If $\nu « \mu$ and $\lambda « \mu$ show that \(\frac{d(\nu + \lambda)}{d\mu} = \frac{d\nu}{d\mu} + \frac{d\lambda}{d\mu} a.e. \mu\)
 If $\nu « \mu « \lambda$, show that \(\frac{d\nu}{d\lambda} = \frac{d\nu}{d\mu} \frac{d\mu}{d\lambda} a.e. \lambda\)
 If $\nu « \mu$ and $\mu « \nu$ show that \(\frac{d\nu}{d\mu} \frac{d\mu}{d\nu} = 1 a.e. \mu\) Solution: To prove the equation
\(\int f\:d\nu=\int f\frac{d\nu}{d\mu}\:d\mu,\)
first observe that this is simply the definition of the RadonNikodym derivative when $f=\chi_E$. From there it is easy to prove for simple functions, then measurable functions.
To prove additivity, note that the RadonNikodym derivative is the unique function satisfying
\((\nu+\lambda)(E)=\int_E\frac{d(\nu+\lambda)}{d\mu}\:d\mu.\)
It is easy to verify that $\frac{d\nu}{d\mu}+\frac{d\lambda}{d\mu}$ satisfies this equation.
To prove the third equation, we combine the idea for the second with the first equation. The function $\frac{d\nu}{d\lambda}$ is the unique function satisfying
\(\nu(E)=\int_E \frac{d\nu}{d\lambda}\:d\lambda.\)
We just need to prove
\(\nu(E)=\int_E \frac{d\nu}{d\mu}\frac{d\mu}{d\lambda}\:d\lambda.\)
By the first part of this problem, the left hand side is
\(\int_E \frac{d\nu}{d\mu}\:d\mu,\)
and this is $\nu(E)$ by definition.
 Axler, 9B.2
 Axler, 9B.6. You will have to look up some definitions in this section.
 Axler, 9B.13.
Homework 7 (Oct 22, 2024)
Problems this week from Axler.

4A.1

4A.7

4A.13

4B.4

4B.6
Homework 6 (due Oct 8)
Problem 1

Let $f$ be integrable over $R$ and $\e > 0$. Establish the three approximation properties:
 There is a simple function $\eta$ on $\R$ which has finite support and $\int_R f  \eta < \e$.
 There is a step function $s$ on $\R$ which vanishes outside a closed, bounded interval and $\int_R f  s < \e$
 There is a continuous function $g$ on $\R$ which vanishes outside a bounded set and $\int_R f  g < \e$.
Solution To approximate integrable functions with simple functions, first assume $f$ is integrable and nonnegative. By definition of the integral, there is a simple function $\psi\leq f$ with $\int f\leq \int\psi+\e$, and hence
\(\int f\psi=\int f\psi=\int f\int \psi<\e.\)
If \(f\) is integrable, we split it into positive and negative parts \(f\_+\) and \(f\_\) and obtain \(\psi\_+\) and \(\psi\_\) as above with \(\e/2\) in place of \(\e\), and set \(\psi=\psi\_++\psi\_\). We have, by the triangle inequality,
\(\int f\psi\leq \int f_+\psi_++\intf_  \psi_<\e/2+\e/2=\e.\)
To approximate integrable functions by step functions, we use simple functions as an intermediate step. If we can approximate a simple function \(\psi\) with a step function \(s\), then using our work above we will have
\(\int fs\leq \int f\psi+\int \psis<\e/2+\e/2=\e.\)
We also make one more reduction and assume that $f$ is compactly supported, as any integrable function can be approximated by a sequence of restrictions to compact sets. To start, we approximate a characteristic function $\alpha\chi_E$ by simple functions, where $E$ is a measurable set contained in the support of $f$, and hence in some interval of finite length. By problem 3.2 on homework 4, we have a step function $s$ defined on that interval which agrees with $\chi_E$ except on a set of measure $<\e/\alpha$. Since $s\alpha\chi_E\leq \alpha$ on our exceptional set, it follows that $\int s\chi_E<\e$. Given $\psi=\sum_{i=1}^n \alpha_i\chi_{E_i}$, let $s_i$ approximate $\alpha_i\chi_{E_i}$ as above, with $\e/n$ in place of $\e$. Let $s=\sum_{i=1}^n s_i$. Then we have
\(\int s\psi\leq\sum_{i=1}^n\int s_i\alpha_i\chi_{E_i}<\e.\)
Finally, to approximate integrable functions by continuous functions, we use the same trick as above and show that we may approximate step functions with continuous functions. To do this, take each interval on which the step function is constant, shrink it a little, and define $g$ by connecting the endpoints linearly (you should be able to fill in the details).

(RiemannLebesgue lemma). Let $f$ be integrable over $(\infty,\infty)$. Show that
\(\lim_{n \to \infty} \int_{\infty}^{\infty} f(x) \cos(nx) ~dx = 0.\)
Hint: Use the previous problem.
Solution
For any step function $s$, we have
\(\int_{\infty}^\infty f(x)\cos(nx)\:dx=\int_{\infty}^\infty s(x)\cos(nx)\:dx+\int_{\infty}^\infty (f(x)s(x))\cos(nx)\:dx.\)
By the triangle inequality and the above problem, it suffices to prove
\(\lim_{n\to\infty}\int_a^b s(x)\cos(nx)\:dx=0,\)
where $s$ is a step function supported on $[a,b]$. Let $I_1,…,I_N$ be intervals such that the left endpoint of $I_1$ is $a$, the right endpoint of $I_N$ is $b$, the right endpoint of $I_j$ is $I_{j+1}$, and $s$ takes the constant value $c_j$ on $I_j$. Then
\(\int_a^b s(x)\cos(nx)\:dx=\sum_{j=1}^N c_j\int_{I_j} \cos(nx)\:dx.\)
It therefore suffices to prove
\(\lim_{n\to\infty}\int_a^b \cos(nx)\:dx=0.\)
This is an easy computation.
Problem 2 (Cantor Set)

Let \(E\) be a nonmeasurable set of finite outer measure. Show that there is a \(G_{\delta}\) set that contains \(E\) for which \(m^*(E) = m^*(G)\) while \(m^*(G \setminus E) > 0\).

Let \(f\) be a continuous function on \(E\). Is it true that \(f^{1}(A)\) is measurable if \(A\) is measurable?

Let \(F\) be the subset of \([0,1]\) constructed in the same manner as the Cantor set except that each of the intervals removed at the \(n\)^{th} step has length \(\alpha 3^{n}\) with \(0 < \alpha < 1\). Show that \(F\) is a closed set, \([0,1] \setminus F\) is dense in \([0,1]\) and \(m(F) = 1  \alpha\). Such a set \(F\) is called a generalized or ‘‘fat’’ Cantor set.

Show that there is an open set of real numbers that has a boundary of positive measure. Hint: Consider the complement of the fat Cantor set.
Problem 3 (Littlewood’s principles)

Suppose \(f\) is a function that is continuous on a closed set \(F\) of real numbers. Show that \(f\) has a continuous extension to all of \(\R\). Hint: Express \(\R \setminus F\) as the union of a countable disjoint collection of open intervals and define f to be linear on the closure of each of these intervals.

Recall that for any measurable function \(f\) on a bounded measurable set \(E\), there is a closed \(F\), and a continuous \(g\) on \(\R\) such that \(f=g\) on \(F\) and \(m(E \setminus F) < \e\). Show that the restriction of \(f\) to \(F\) is continuous. Does \(f\) have to be continuous on \(E\)?

Suppose \(F \subset \R\) is such that every continuous function from \(F\) to \(\R\) can be extended to a continuous function from \(\R \to \R\). Prove that \(F\) is a closed subset of \(\R\).
Homework 5 (due Oct 1)
All problems are on $\R$.
Problem 1 (Measurability)

Suppose $f$ has a measurable domain and is continuous except at a finite number of points. Is $f$ necessarily measurable?
Solution: Yes, such a function must be measurable. Let $S={x_1,…,x_N}$ be the set of discontinuities. For all $n\in\mathbb{N}$, define $f_n(x)$ to be $f(x)$ except on intervals $(a_n,b_n)$ of length $o(n)$ centered at each $x_i$. For $x\in (a_n,b_n)$, define
\(f_n(x)=f(a_n)+\frac{f(b_n)f(a_n)}{b_na_n}(xa_n).\)
Then $f_n$ is continuous, hence measurable, and $f_n\to f$ pointwise everywhere except possibly on $S$. Hence, $f$ is measurable.

Suppose $f$ is defined on a measurable set $E$ and $f^{1}((c,\infty))$ is measurable for every rational $c$. Is $f$ necessarily measurable?
Solution: Yes. For every $x\in\mathbb{R}$, we have
\((x,\infty)=\bigcup_{\substack{c\in\mathbb{Q} \\ c>x}} (c,\infty).\)

Let ${f_n}$ be a sequence of measurable functions defined on a measurable set $E$. Define $E_0$ to be the set of points $x \in E$ such that ${f_n(x)}$ converges. Is $E_0$ measurable?
Solution: A sequence of real numbers converges if and only if it is Cauchy; therefore,
\(E_0=\bigcap_{\e>0}\bigcup_N\bigcap_{n,m>N}\{x\in E:f_n(x)f_m(x)<\e\}.\)
By taking $\e$ to be rational, we can make each union and intersection countable. Because $f_n$ and $f_m$ are measurable, $f_nf_m$ is measurable and hence the sets ${x\in E:f_n(x)f_m(x)<\e}$ are all measurable. Therefore, $E_0$ is measurable.

Let $f$ be a measurable function and $g$ be a $11$ function from $\R$ onto $\R$ which has a Lipschitz inverse. A function is said to be Lipschitz if there is $c > 0$ such that
\(f(x)  f(y) < cx  y\)
for all $x,y \in \R$. Show that the composition $f \circ g$ is measurable.
Solution: By proposition 2 in section 3.1, it is enough to prove $(f\circ g)^{1}(O)$ is measurable for open sets $O$. Since $(f\circ g)^{1}(O)=g^{1}(f^{1}(O))$, and since $f$ is measurable, it is enough to prove that $g^{1}(E)$ is measurable for measurable sets $E$. Since the image of a union is the union of the images, and since a set is measurable if and only if it is the union of an $F_\sigma$ set and a measure zero set, it is enough to prove that $g^{1}$ preserve both classes.
To prove $g^{1}$ maps $F_\sigma$ sets to $F_\sigma$ sets, first observe that $g^{1}$ is continuous and bijective, hence strictly monotone. It follows that $g$ is strictly monotone and bijective, hence continuous. So $g^{1}$ maps closed sets to closed sets, and therefore $F_\sigma$ sets to $F_\sigma$ sets.
To prove $g^{1}$ maps measure zero sets to measure zero sets, let $E$ have measure zero. Then for $\e>0$, $E$ can be covered by intervals ${I_n}$ with $\sum_n I_n<\frac{\e}{c}$. Since $g^{1}$ is Lipschitz, $g^{1}(I_n)$ is contained in an interval of size $\leq cI_n$, so $g^{1}(E)$ has such a covering with total length $<\e$.

Let $I$ be an interval and $f \colon I \to \R$ be increasing. Show that $f$ is measurable by first showing that, for each natural number $n$, the strictly increasing function $g(x) = f(x) + x/n$ is measurable and then take pointwise limits.
Problem 2 (Convergence)

Show that the conclusion of Egoroff’s theorem can fail if we drop the assumption that the domain has finite measure. How did the continuity of measure show up in Egoroff?
Solution Let \(f_n=1_{[n,\infty)}\) on \(\R\) with Lebesgue measure. \(f_n \to 0\). Suppose for every \(\e > 0\), there is a set \(F\) such that \(m(F^c) < \e\) and \(f_n \to 0\) uniformly on \(F\). However, \(m([n,\infty) \cap F^c) < \e\), which means \(m([n,\infty) \cap F) = \infty\) and \(\vert f_n  0\vert = 1\) on this set. This is a contradiction.

Give an example of a decreasing sequence ${f_n}$, $f_n \geq 0$ and $f_n \to f$ such that the analog of the MCT does not hold.
Solution Let \(f\_n=1_{[n,\infty)}\).

Suppose \(f_n \to f\) uniformly for almost every $x \in \R$. Is it true that
\(\int f_n = \int f\)
If true, prove it, if not, give a counterexample. What happens on a bounded measurable subset of $\R$?
Solution
Consider \(f_n = \text{sign}(x) \vert x\vert ^{1/2} 1_{[n,n] \setminus 0}\). By symmetry, \(\int f_n = 0\). \(f_n \to \text{sign}(x) \vert x\vert ^{1/2} 1_{\R \setminus 0}\) which is not integrable. Note that this convergence is uniform: for every \(\e > 0\), pick an \(n\) so large such that \(\vert n\vert ^{1/2} < \e\). Then \(\vert f_n(x)  f(x)\vert \leq \vert n\vert ^{1/2} 1_{ \R \setminus [n,n]} \leq \e\).

Let \(g\) be a nonnegative integrable function over \(E\) and suppose \(\{f\_n\}\) is a sequence of measurable functions on \(E\) such that for each \(n\), \(\\vert f\_n\\vert \leq g\) a.e. Show that
\(\int_E \liminf f_n \leq \liminf \int_E f_n \leq \limsup \int_E f_n \leq \int_E \limsup f_n .\)
Solution
Consider $g  f_n$ and $g+ f_n$, both nonnegative functions. By Fatou \(\begin{align} \int \liminf g  f_n & \leq \liminf \int g  f_n \\ \int g  \int \limsup f_n & \leq \int g  \limsup \int f_n \\ \limsup \int f_n & \leq \int \limsup f_n \end{align}\) using linearity of the integral and standard properties of liminfs and limsups. Apply Fatou to $g + f_n$ to get the inequality for the liminfs.
The version of Fatou proved in class was the following: Let $f_n \to f$ a.e., and $f_n \geq 0$. Then
\(\int \liminf f_n \leq \liminf \int f_n\)
If we dont have \(f_n \to f\), let \(g = \liminf f_n\) and \(g_n = \inf_{k \geq n} f_k\). Then \(g_n\) increases monotonely to \(g\). Then
\(\int \lim_{n} g_n \leq \liminf \int g_n = \inf_m \sup_{n \geq m} \int \inf_{k \geq n} f_k \leq \inf_m \sup_{n \geq m} \int f_n.\)
which gives the version we have used here. The first inequality in the chain was Fatou, the second is just the definition of \(g_n\) and the third uses \(\inf_{k \geq n} f_k \leq f_n\).
Homework 4 (due Sep 24, 2024)
Problem 1
This problem is on \(\R\).

Show that a set \(E\) is measurable iff for each \(\e > 0\), there is a closed \(F\) and an open \(O\) such that \(F \subset E \subset O\) and \(m^*(O \setminus F) < \e\).
Solution: Use the characterization of measurability given in Section 2.4, Theorem 11 in the text. If $E$ is measurable, for any $\e>0$ there is an open set $O\supset E$ with $m^*(O\setminus E)<\frac{\e}{2}$ and a closed set $F\subset E$ with $m^*(E\setminus F)<\frac{\e}{2}$. Using subadditivity and the fact that
\(O\setminus F=(O\setminus E)\cup (E\setminus F),\)
we conclude that $m^*(O\setminus F)<\e$. Conversely, suppose such sets $O$ and $F$ exist for any $\e>0$. By monotonicity, $m^*(O\setminus E)\leq m^*(O\setminus F)<\e$, so $E$ is measurable by part (i) of the theorem.

Let \(E\) have finite outer measure. Show that \(E\) is measurable iff for each bounded open interval \((a,b)\)
\(b  a = m^*((a,b) \cap E) + m^*((a,b) \cap E^c)\)
Solution: One direction follows directly from the definition. For the other, suppose the given equation holds for any $a<b$. We want to prove $E$ is measurable. Fix $\e>0$ and cover $E$ by intervals $I_n$ of length $\ell_n$, which we can take to be disjoint, so that $\sum_n \ell_n<m^*(E)+\e$. By assumption, we have
\(\ell_n=m^*(E\cap I_n)+m^*(I_n\setminus E).\)
Let $O=\cup_n I_n$. Then
\(m^*(O\setminus E)\leq \sum_n m^*(I_n\setminus E)=\sum_n(\ell_nm^*(E\cap I_n))<m^*(E)+\e\sum_n m^*(E\cap I_n)\leq \e.\)
So, $E$ is measurable.

For any set $A$ define $m^{**}(A) \in [0,\infty]$ by
\(m^{**}(A) = \inf \{ m^*(O) \colon O \supseteq A, O \text{ open} \}\)
How is \(m^{**}\) related to $m^*$? Prove your answer.
Solution: We claim $m^{**}=m^*$. This follows from the fact that any open set can be decomposed into a disjoint union of open intervals, so the definition of $m^{**}$ can be rewritten
\(m^{**}(A)=\inf\{\sum_n I_n:A\subset\cup_n I_n\},\)
where the \(I\_n\) are open intervals and \(\vert I\_n\vert\) is the length of \(I\_n\). This is the definition of outer measure.
Problem 2

Show that continuity of measure together with finite additivity of measure implies countable additivity of measure.
Solution: Let ${E_n:n\in\mathbb{N}}$ be a countable collection of disjoint measurable sets, and let $A_N=E_1\cup\cdots\cup E_N$. Continuity of measure and finite additivity imply
\(m\left(\bigcup_{n=1}^\infty E_n\right)=m\left(\bigcup_{N=1}^\infty A_N\right)=\lim_{N\to\infty} m(A_N)=\lim_{N\to\infty}\sum_{n=1}^N m(E_n)=\sum_{n=1}^\infty m(E_n).\)

Let ${E_k}_{k=1}^\infty$ be a sequence of measurable sets of bounded measure. How are
\(\limsup_{k \to \infty} m(E_k) \AND m \left( \limsup_{k \to \infty} E_k \right)\)
related? How about
\(\liminf_{k \to \infty} m(E_k) \AND m \left( \liminf_{k \to \infty} E_k \right)\)
Suppose
\(\begin{align*} m ( \limsup_{k \to \infty} E_k \Delta \liminf_{k \to \infty} E_k ) & = 0 \\ m ( \limsup_{k \to \infty} E_k ) & < \infty \end{align*}\)
where $\Delta$ is the symmetric difference operator.
Does
\(\lim_{k \to \infty} m(E_k)\)
exist?
Solution:
For the first part, considering the limit superior, there is no relationship between the two sides. More precisely, for any $\alpha,\beta\in\mathbb{R}_{\geq 0}$ there is a collection ${E_k}$ with $\limsup_{k\to\infty}m(E_k)=\alpha$ and $m(\limsup_{k\to\infty} E_k)=\beta$. \
To prove this, first let $0<\alpha<\beta$ and choose $N$ to be the largest natural number satisfying $N\alpha <\beta$ with $\gamma<\alpha$. Let $E_k$ run through the sets $[0,\alpha],[\alpha,2\alpha],…,[(N1)\alpha,N\alpha],[N\alpha, \beta]$, then repeat in a cycle of length $N+1$. Then $\limsup_k E_k=[0,\beta]$ has measure $\beta$, as claimed. The sequence $m(E_k)$ runs through values $\alpha$ and $\betaN\alpha$ each infinitely often. Since $\betaN\alpha\leq \alpha$, this means $\alpha=\limsup_k m(E_k)$. \
Next, suppose $\alpha=0$ and $\beta>0$. Let $I_{j,N}=[\frac{\beta(j1)}{N},\frac{\beta j}{N}]$ for $N\in\mathbb{N},1\leq j\leq N$. Let $E_k$ run through $I_{j,N}$ starting by exhausting $N=1$, then $N=2$, and so on. Then $\limsup_k m(E_k)=0$, but $\limsup_k E_k=[0,\beta]$ so $m(\limsup_k E_k)=\beta$. \
Finally, suppose $\beta\leq \alpha$. Let $E_k$ be the set $[0,\beta]$ whenever $k$ is even, and when $k$ is odd, say $k=2j+1$, let $E_{2j+1}=[\beta+j\alpha,\beta+(j+1)\alpha]$. The sequence $m(E_k)$ takes values $\beta$ and $\alpha$, each infinitely often, so $\limsup_k m(E_k)=\alpha$. Since $\limsup_k E_k=[0,\beta]$, we have $m(\limsup_k) E_k=\beta$. \
For the second part, concerning the limit inferior, we first observe that $\cap_{k>N}E_k$ is an ascending chain in $N$. Therefore, by continuity of measure,
\(m(\liminf_k E_k)=m\left(\bigcup_N\bigcap_{k>N}E_k\right)=\lim_{N\to\infty}m\left(\bigcap_{k>N}E_k\right).\)
For any $k_0>N$, we have $m\left(\bigcap_{k>N}E_k\right)\leq m(E_{k_0})$ by monotonicity. Therefore, $m\left(\bigcap_{k>N}E_k\right)\leq \inf_{k>N} m(E_k)$. Pairing this with the line above gives
\(m(\liminf_k E_k)\leq \lim_{N\to\infty}\inf_{k>N} m(E_k)=\liminf_k m(E_k).\)
The inequality can be strict. In fact, for any $\alpha<\beta$, there exists a sequence ${E_k}$ with $m(\liminf_k E_k)=\alpha$ and $\liminf_k m(E_k)=\beta$. Let the sequence ${E_k}$ alternate between the intervals $[0,\beta]$ and $[\beta\alpha,2\beta\alpha]$. Then $\liminf_k E_k$ is the intersection of the intervals, which is the interval $[\beta\alpha,\beta]$ of length $\alpha$. On the other hand, each set satisfies $m(E_k)=\beta$. \
Finally, to answer the last part of the question, the condition that the the symmetric difference of the limits superior and inferior has measure zero is not enough to guarantee that $\lim_k m(E_k)$ exists. As an example, let $E_k$ be an interval centered at $k$ with length $1/100$ if $k$ is even and length $1/1000$ if $k$ is odd. Then $m(E_k)$ alternates between two distinct numbers and therefore the limit does not exist, but the limits superior and inferior are both empty, hence the hypotheses are trivially met.
Problem 3

Let $f$ be a continuous function on $\R$ such that $f(x+y) = f(x) + f(y)$. Show that $f$ must be linear.
Solution Note that $f(1) = f(1) + f(0)$ which implies that $f(0) = 0$. Then $f(n) = n f(1)$ by induction. Similarly \(f(1/n) = f(1)/n\), and repeating this \(f(p/q) = p/q f(1)\) for every rational. Since it’s continuous, \(f(x) = x f(1)\) by rationals converging to $x$.

Show that Lebesgue measure is the unique, nontrivial (up to scaling) $\sigma$finite translation invariant measure on $\R$ with the Borel or Lebesgue $\sigma$algebra.
Solution Let \(m\) be such a translation invariant measure, which is the property that \(m(A) = m(A + x)\) for every measurable \(A\) and \(x \in \R\).
Set \(f(x) = m((0,x])\), and notice that
\(f(x+y) = m((0,x+y]) = m((0,x]) + m((x,x+y]) = m((0,x]) + m((0,y])\)
using the translation invariance property.
The following uses BaireCategory, thanks to June for pointing out the gap. This is Corollary 4, Chapter 10 in Royden: let \(X\) be a complete metric space, and \(\{F_n\}_{n=1}^\infty\) be a closed collection of sets. If \(\cup_n F_n\) has a nonempty interior, then at least one of $F_n$ has nonempty interior.
Next we show \(m((a,b))\) is finite for some nontrivial, bounded, open interval. Note that \(X = \cup_i A_i\) where \(A_i\) is a union of finitemeasure sets. The interior of any set is open, which means it contains an interval if it is nonempty. If we show that any one of the \(A_i\) has a nonempty interior, then there is an interval inside \(I \subset A^o\) such that \(m(I) \leq m(A) < \infty\).
So for the sake of contradiction, assume that all the \(A_i\) have empty interior, which means all the \(A_i\) are closed and \(\cup_i A_i = \R\). This contradicts the Baire category theorem. So there is an interval of the form $m((a,b)) < \infty$.
Next, we will (implicitly) use the nontriviality of \(m\) (that it is not identically \(+\infty\) on all nonempty sets) to show that \(m((a,b]) = m((a,b))\). Suppose there is an \(a \in \R\) such that \(m(a) > 0\). Then, by translation invariance
\(m(\{1/n,\ldots,11/n\}) \leq m(0,1) < \infty.\)
Taking \(n\) to \(\infty\) gives us a contradiction. Therefore
\(m((a,b]) = m((a,b)) + m(b) = m((a,b)).\)
One can translate the interval \((a,b]\) to the origin to an interval of the form \(m((0,x])\). By (ascending) continuity of measure, \(m((0,p/q))\) is finite for some rational \(p/q\). Then apply additivity to show \(m((0,1])\) is finite. Thus, both descending and ascending continuity of measure can be applied to show \(f(x)\) is continuous and additive, and by part \(1\) it is linear. This shows \(m((0,x]) = C x\) for some constant \(C\). Thus, $m$ is a $\sigma$finite premeasure that agrees with Lebesgue measure on a semiring and we are done.
Homework 3 (due Sep 17, 2024)
Problem 1
In this problem we will build a measure called Jordan measure and connect it to the Riemann integral. The main difference between the Jordan measure and the Lebesgue measure is that the Lebesgue measure is countably additive whereas the Jordan measure is finitely additive. The Jordan measure is intimately connected with the Riemann integral, and so morally, the only difference between the Riemann and Lebesgue integrals is changing “finite” to “countable” in the definition.
We will define the Jordan measure on $\R^d$ instead of the real line for a little extra difficulty.
Box A (finite) box on $\R^d$ is a Cartesian product:
\(B = I_1 \times I_2 \times \cdots \times I_d\)
where $I_k$ are bounded intervals on the real line.
Elementary sets: An elementary subset of $\R$ is any finite union of boxes.
Volume of a box A box $B$ has volume
\(B := I_1 \times I_2 \times \cdots \times I_d\)
where $I_k$ represents the length of the interval.
Finite disjoint unions

Show that if $E$ is an elementary set, then it can be expressed as a finite disjoint union of boxes.
Hint: First show it for intervals and then consider a proof by induction.

Show that $E$ is independent of the boxes that appear in the disjoint union. That is, if
\(E = B_1 \sqcup \cdots \sqcup B_n = B_1' \sqcup \cdots \sqcup B_m'\)
then
\(E = \sum_{i=1}^n B_i = \sum_{i=1}^m B_i'\) Hint: We have proved similar theorems for the Lebesgue measure. The general strategy is that if you have two different ways to express $E$ as a disjoint union, then consider the “common refinement” of the two sets of boxes ${B_i}$ and ${B_i’}$.

Then we may define
\(m(E) = \sum_{i=1}^n B_i\)
to be the measure of an elementary set. For an elementary set $E \subset \R^d$, $m(E)$ show that it posseses
 Monotonicity
 Finite additivity for disjoint sets and hence finite subadditivity
 Translation invariance
Jordan measure
For bounded subsets of $\R^d$, define Jordan inner measure and Jordan outer measure as follows:
\(\begin{align} J^*(E) & = \inf_{A \supset E, A \text{ elementary}} m(A)\\ J_*(E) & = \sup_{A \subset E, A \text{ elementary}} m(A) \end{align}\)
If $J^*(E) = J_*(E)$, then $E$ is Jordan measurable and we call their common value $m_J(E)$.
 Show that $E$ and $\overline{E}$ have the same Jordan outer measure.
 Show that $E$ and ${E}^{\circ}$ have the same Jordan inner measure. Recall that we said that a point $x \in E \subset R$ is an interior point if there is small open interval containing $x$ that is a subset of $E$. In $\R^d$, interior points are defined similarly using open boxes instead of open intervals.
 Show that the topological boundary of $\partial E = \overline{E} \setminus E^\circ$ has zero Jordan measure iff $E$ is Jordan measurable.
 If $E = [0,1] \cap \Q$, show that $J^*(E) = 1$ but $J_*(E) = 0$.
Caratheodory type property

Let $F$ be a box and $E$ be a bounded set. Then show that
\(J^*(E) = J^*(E \cap F) + J^*(E \setminus F)\)
Hint: Follow the strategy outlined in Royden that shows that intervals are Lebesgue measurable.
Connection to Riemann integral
Let $[a,b]$ be an interval and $f,g$ be Riemann integrable. Show that

(Linearity) For any real number $c$ show that $cf$ and $f + g$ are (R)iemann integrable. Then, show that $(R)\int c f = c (R) \int f$ and $(R) \int (f + g) = (R)\int f + (R)\int g$.

(Monotonicity) If $f \leq g$ pointwise, then $(R) \int f \leq (R) \int g$.

(Indicator) If $E$ is Jordan measurable subset of $[a,b]$ show that $(R)\int 1_E = m_J(E)$.

(Uniqueness) Show that the functional $f \mapsto \int f$ on the space of (R)iemann integrable functions that satisfies the three properties above is unique.
Homework 2 (due Sep 10, 2024)
Problem 1 (More on real numbers)

Is the empty set open or closed? Are the rational numbers open or closed? Justify.
Solution: The empty set is both open and closed, since it satisfies both definitions vacuously. The rationals are neither; it is not open because there is no interval containing only rationals ($\mathbb{R}\setminus\mathbb{Q}$ is dense) and it is not closed because every real number is a limit of rationals ($\mathbb{Q}$ is dense).

A point is called an accumulation point of a set $E$ provided it is a point of closure of $E \setminus {x }$. Show that the set of accumulation points of $E$ (called $E’$) is closed and that $\overline{E} = E \cup E’$.
Solution: To show $E’$ is closed, we must prove $E’=\overline{E’}$. Suppose $x\in \overline{E’}$, so every interval $(xr,x+r)$ contains an element $y\in E’$. We may assume $y\neq x$, since otherwise we have $x\in E’$ which is the goal. If $t$ is sufficiently small so that $(yt,y+t)\subset (xr,x+r)$ and $x\notin (yt,y+t)$, then since $y\in E’$ there is a $z\in (yt,y+t)\cap E\subset (xr,x+r)\cap E$. Since we cannot have $z=x$ by construction, $x\in E’$, and it follows that $E’$ is closed.\
We next prove $\overline{E}=E\cup E’$. One $\supset$ direction is obvious, so we just prove the other. Let $x\in\overline{E}$. If $x\in E$, then $x\in E\cup E’$ and we are done. If not, by definition of closure every open interval containing $x$ must meet $E$, and by assumption must meet $E$ at a point other than $x$, and therefore $x\in E’$.

A point is called isolated if there is an $r > 0$ for which $(xr,x+r) \cap E = {x}$. Show that if all points in $E$ are isolated, it is countable.
Solution: For each $x\in E$, let $I_x$ be an open interval containing $x$ satisfying $I_x\cap E={x}$. Without loss of generality, we may assume each $I_x$ has rational endpoints. This gives a map $E\mapsto \mathbb{Q}^2$ giving by mapping $x$ to the endpoints of the interval $I_x$. This map must be injective, since the endpoints of an interval characterize the interval and clearly $I_x=I_y$ implies $x=y$. Therefore, $E$ is countable.

Show that the HeineBorel theorem is equivalent to the completeness axiom.
Solution: We first prove as a lemma that compact sets in $\mathbb{R}$ contain a largest element. Let $K$ be compact, and suppose it does not have a maximum element. Then ${(\infty,x):x\in K}$ is an open cover of $K$. By extracting a finite subcover, we see that $K$ has a largest element, contradicting our assumption.
The HeineBorel Theorem is proved in the text (using completeness), so it suffices to prove that HeineBorel implies completeness. Suppose $E$ is nonempty and bounded above; we must prove it has a least upper bound. Without loss of generality we may assume $E$ is bounded below as well as above (since otherwise we can replace $E$ with $E\cap [x_0,\infty)$ for some choice of $x_0$ without changing any of our arguments about upper bounds). Then $\overline{E}$ is compact by HeineBorel, and hence has a maximum element $\alpha$. Since $E\subset\overline{E}$ and $\alpha=\max(\overline{E})$, we know $\alpha$ is an upper bound for $E$. We claim it is the least upper bound. If $\alpha\in E$, then we are done, as a maximum is also a least upper bound. If $\alpha\in \overline{E}\setminus E$, then by a previous exercise it is an accumulation point, and hence $\alpha\e$ cannot be an upper bound for any positive $\e$.
Problem 2 (Riemann Integral)

Consider the function \(f(x) = 1/\sqrt{x}\) on the interval $[a,1]$. Show, by computing the Upper and Lower Riemann integrals, that for \(a \in (0,1)\), \(\int f(x) dx = 2(1  \sqrt{a})\).

Axler 1A, problem 11

Axler 1A, problem 14
Problem 3 (Lebesgue outer measure)
Let \(m\) be the set function on halfopen intervals on \(\R\) given by \(m((a,b]) = b  a\). Let \(m^*\) be the Caratheodory outer measure obtained from \(m\).

Show that for any $B$, we must have \(m^*(B) = m^*(B \setminus \{a\}).\)
Solution: Since $B\setminus{a}\subset B$, we have $m^*(B\setminus{a})\leq m^*(B)$. For any covering of $B\setminus{a}$ by open intervals, we can add the interval centered at $a$ of length $\e$ and obtain a covering of $B$ with $\e$ extra length. Therefore, for any $\e>0$, we have $m^*(B)\leq m^*(B\setminus{a})+\e$. The equation follows.

Prove that the outer measure of an unbounded interval is $\infty$.
Solution: See proof of proposition 1, section 2.2 of text.

Let $A$ be a countable set. Find $m^*(A)$.
Solution: It is proved in the text that the outer measure of an interval is its length. On the other hand, the outer measure of a countable set is zero.

Let $\{I_k\}_{k = 1}^n$ be a finite cover by open intervals of $\Q \cap [0,1]$. Show that \(\sum_{k=1}^n m^*(I_k) \geq 1\).
Solution: Let $I_1,…,I_N$ cover $\mathbb{Q}\cap [0,1]$. We claim that the intervals must in fact cover all but finitely many points of $[0,1]$, which implies the result by what we know about outer measure of intervals. To prove the claim, suppose $x\in[0,1]$ is not in any of the intervals. In particular, we must have $x\neq 0$. Let ${q_n}$ be a monotone increasing sequence of rationals with limit $x$. By the pigeonhole principle, there must be a single interval $I_n$ containing an infinite subsequence, which must still converge to $x$. In order to have $x\notin I_n$, it must be the case that $x$ is the right endpoint of $I_n$. Since there are only finitely many intervals, there are only finitely many such points.

Show that if a set $E$ has positive outer measure, then there is a bounded subset of $E$ that has positive outer measure.
Solution: If $E\cap [N,N]$ has outer measure zero for all $n$, then each is measurable and continuity of measure implies $m(E)=0$. Therefore, if $E$ has positive outer measure, one of the sets $E\cap [N,N]$ must have positive outer measure.

Show that if $E$ has finite measure and $\e > 0$, then $E$ is the disjoint union of a finite number of measurable sets, each of which has measure at most $\e$.
Solution: Let $f(x)=m(E\cap [x,x])$. Then $f$ is a continuous (by continuity of measure) monotone nondecreasing function with $f(0)=0$ and $\lim_{x\to\infty} f(x)=m(E)$. This means for any integer $k$ with $k\e<m(E)$, there is an $x_k$ with $f(x_k)=k\e$. Let $E_k=E\cap (x_{k1},x_k]$. This (together with $E\cap (x_{k_\text{max}},\infty)$) is a disjoint decomposition of $E$ into $\approx \e^{1}m(E)$ subsets each of which has measure $\e$ (except the last one, which may have measure strictly less).
Homework 1 (due Sep 3, 2019)

Using the field and positivity axioms for the real numbers, show that if \(a\) is positive and \(b\) is negative, then \(a \cdot b\) is negative.
Solution: By definition, in order to prove $ab$ is negative we must prove that $(ab)$ is positive. Since $a$ and $b$ are positive, $a(b)$ is positive. So, it suffices to prove that $(ab)=a(b)$. To do this, it suffices to prove $ab+a(b)=0$. By distributivity, $ab+a(b)=a(b+(b))=a\cdot 0=0$.

The completeness axiom says that every set bounded above has a least upper bound or supremum. Show that a set bounded below has an infimum or greatest lower bound.
Solution: Let $E\subset\mathbb{R}$ be bounded below. Then $E:={x:x\in E}$ is bounded above, and hence has a supremum $\alpha:=\sup E$. We claim $\alpha$ is the greatest lower bound of $E$. To prove it is a lower bound, let $x\in E$. Then $x\in E$, so $x\leq \alpha$, so $x\geq \alpha$. To prove it is the greatest lower bound, let $\beta>\alpha$. Then $\beta<\alpha$, so there exists $y\in E$ with $y>\beta$. It follows that $y\in E$, and $y<\beta$. So, $\alpha$ is the greatest lower bound of $E$.

What can you say about a field where the multiplicative identity \(1\) is equal to the additive identity \(0\) (Royden calls this the nontriviality assumption)?
Solution: Let $F$ satisfy the field properties except that it satsifies $1=0$. Then $F$ has only one element, namely the additive/multiplicative identity. To prove this, we first note it is easy to prove the additive identity 0 satisfies $x\cdot 0=0$ for every $x$; this is because $x\cdot 0=x\cdot (0+0)=x\cdot 0+x\cdot 0$. Adding $(x\cdot 0)$ to each side gives $x\cdot 0=0$. On the other hand, if the additive identity is also the multiplicative identity, then $x\cdot 0=x$. Therefore, $x=0$ for every $x$.

Let \(c\) be a positive real number. Show that there is a unique positive square root \(x > 0\) for \(c\) . Royden has a bunch of hints that guide you through this problem
Solution: To prove existence, let
\(\alpha=\sup\{x:x^2<c\}.\)
The set ${x:x^2<c}$ is nonempty (as $0^2=0<c$) and bounded above (by $\max(c,1)$, for example), so $\alpha$ is well defined. We claim $\alpha^2=c$, which we prove by contradiction.
First suppose $\alpha^2<c$. We have, for any $\e<1$,
\((\alpha+\e)^2=\alpha^2+2\alpha\e+\e^2<\alpha^2+\e(2\alpha+1).\)
Let $M=2\cdot\max(1,c)+1$, so that
\((\alpha+\e)^2<\alpha^2+M\e.\)
If $\e<\frac{c\alpha^2}{M}$, then $\alpha+\e\in{x:x^2<c}$, contradicting the assumption that $\alpha$ is an upper bound.
Now, suppose $\alpha^2>c$. Then a similar argument shows $(\alpha\e)^2>\alpha^2M’\e$. Letting $\e$ be small enough, we see that $(\alpha\e)^2>c$ and therefore $\alpha\e$ is an upper bound for ${x:x^2<c}$, contradicting the assumption that $\alpha$ is the least upper bound.
To show uniqueness, we use the fact that $xy=0$ implies $x=0$ or $y=0$ (to prove this, suppose $x\neq 0$ and observe $y=x^{1}xy=x^{1}\cdot 0=0$) and the observation (which follows from repeated use of the distributive and commutative properties) that $x^2y^2=(x+y)(xy)$. Therefore, if $x$ and $y$ are both positive square roots of $c$, then either $x+y=0$ or $xy=0$. The first is impossible if $x$ and $y$ are both positive, and the second implies $x=y$.

Show that the rationals are dense in \(\R\).
Solution: We start by proving the following lemma: if $x$ and $y$ are numbers with $yx>1$, there is an integer $m$ satisfying $x<m<y$. This follows from the well ordering principle. First assume $x$ and $y$ are positive, and let $m$ be the smallest natural number satisfying $x<m$. If $m\geq y$, then $m1\geq y1>x$. This is a contradiction, since $x$ is assumed to be positive and therefore $m1$ is a smaller natural number than $m$ satisfying $m1>x$. So, $x<m<y$ as claimed. If $x$ and $y$ are negative, use the same argument with $y$ in place of $x$ and $x$ in place of $y$. If one is positive and the other is negative, $m=0$ works.\ Let $a<b$. We must prove that there is a rational number $q$ with $a<q<b$. Since $ba$ is positive, it follows from the Archimedian property that there is a natural number $n$ with $n(ba)>1$. Since $nbna>1$, there is an integer $m$ satisfying $na<m<nb$, or $a<\frac{m}{n}<b$.

Suppose we are given a function \(f\) from \(\N\) onto a set \(B\). Define an equivalence relation on \(\N\) by \(x \sim y\) if \(f(x) = f(y)\). Without using the axiom of choice, explictly construct a set \(E \subset \N\) that consists of one member of each equivalence class.
Solution: For each $b\in B$, the set $f^{1}(b)$ is a nonempty set of natural numbers, and therefore has a smallest element $g(b)$. This defines a function $g:B\to\mathbb{N}$; let $E$ be the image of this function.

Show that the rational numbers are countable.
Solution: See the proof of Corollary 4 in the text for the structure of the proof. It remains to show the map $g(p/q)=(p+q)^2+q$ is injective, where $p/q$ is assumed to be written in lowest terms. Let $t=p+q,t’=p’+q’$ so that $g(p/q)=t^2+q$ and $g(p’/q’)=t’^2+q’$. If $t’>t$, then
\(t^2+q\leq t^2+t=t(t+1)< (t+1)^2\leq t'^2\leq t'^2+q'.\)
Since there is a strict inequality in the chain, we do not have $g(p/q)=g(p’/q’)$. The same argument applies if $t’<t$, so we must have $t’=t$. Since $g(p/q)=g(p’/q’)$ means $t^2+q=t’^2+q’$, this implies $q=q’$ and therefore $p=p’$.

Show that the real numbers are uncountable using the Cantor diagonal argument (you may read about it on wikipedia and write it down in your own words).
Solution: Because any subset of a countable set is countable, it suffices to choose a subset of $\mathbb{R}$ and prove it is uncountable. We will use the half open interval $[0,1)$ for convenience. Any element $x\in [0,1)$ has a decimal representation
\(x=\sum_{n=1}^\infty a_n10^{n},\)
with each $a_n$ an integer from 0 to 9, and this representation is unique if we make the convention that the sequence ${a_n}$ cannot end in an infinite string of 9’s. Suppose the set $[0,1)$ is countable, and fix an enumeration $[0,1)={x_1,x_2,…}$. Let $x_i$ have decimal representation
\(x_i=\sum_{n=1}^\infty a_{n,i}10^{n}\)
and define
\(x=\sum_{n=1}^\infty b_n10^{n},\)
where $b_n$ is chosen to be $1$ if $a_{n,n}=0$, and 0 otherwise. Since ${b_n}$ is a sequence of integers between 0 and 9 which does not end in an infinite string of 9’s, this is a valid decimal representation of some $x\in [0,1)$. By construction, $b_n\neq a_{n,n}$, which means that $x$ cannot be $x_i$ for any $i$, since $x$ disagrees with $x_i$ at the $i$th decimal digit. This contradicts the assumption that we have enumerated the set $[0,1)$, so that set must be uncountable.