Please fully type up both your question(s) as well as your solution(s).

## June 2020 Prelim

### Problem 1 (Jackson):
**Let $f$ be an integrable function. Construct a sequence $\{f_n\}$ of bounded measureable functions on $\mathbb{R}^d$ with compact support such that:
$$\int_{\mathbb{R}^d} |f_n - f|dx \to 0 \text{ as } n \to \infty$$**

Define a function on $B_n$, where $B_n$ is a closed ball in $\mathbb{R}^d$ centered at the origin of radius $n$. The function
$$h_n(x)=\begin{cases}
n & f(x)> n , x \in B_n\\
f(x) & -n \leq f(x)\leq n, x \in B_n\\
-n & f(x) < - n, x \in B_n\\
0 & \text{otherwise}
\end{cases} $$ gives us a function that has all values of $|f(x)|< n$ within a ball of radius $n$ around the origin. Since $B_n$ is a closed ball, $h_n$ has compact support for every $n$.This function is measurable since it is either $n$, or $f(x)$ on a measurable set since $f(x)$ is measurable. The function $h_n$ is bounded since $|h_n(x)| \leq n$ for all and $x$. Consider the sequence $\{h_n\}^\infty_{n=1}$. This sequence converges to $f$, as for any $\varepsilon$ and $x$, pick $N$ larger than max$\{|x|, f(x)\}$.

Now we can apply the DCT. We can split this up into $f^-$ and $f^+$, and apply the DCT to $h_n^+$ and $h_n^-$. For either $h_n^+$ or $h_n^-$, $|h_n(x)| \leq |f(x)|$ for all $x$ so $f^+$ or $f^-$ can play the role of our fixed integrable function. Then, because $h_n(x) \rightarrow f(x)$ as $n\rightarrow \infty$, $$ \int |f-h_n(x)| \rightarrow 0 \text{ as } n \rightarrow \infty $$

### Problem 2 (Emma & Erin):
**Let $\{E_k\}_{k=1}^\infty$ be a sequence of measurable subsets of $\mathbb{R}^d$ with
$m(E_k) \leq \frac{1}{k^2}$ for any $k \geq 1$.
Let $E=\{x \in \mathbb{R}^d : x \in E_k \textnormal{ for infinitely many }k\}$.**

**a) E is measurable**

To begin, we claim that $E=\bigcap_{n=1}^{\infty} \bigcup_{k=n}^\infty E_k$.

$\subseteq$:
Let $x \in E$. Then, $x \in \bigcup_{k=n}^\infty E_k$ for all n, as $x$ is in infinitely many $E_k$. Therefore, $x \in \bigcap_{n=1}^{\infty} \bigcup_{k=n}^\infty E_k$.

$\supseteq$:
Let $x \in \bigcap_{n=1}^{\infty} \bigcup_{k=n}^\infty E_k$. Then, for all n, $x \in \bigcup_{k=n}^\infty E_k$. Hence, $x$ is in infinitely many $E_k$, and therefore in $E$.

Then $E$ is a countable intersection of a countable union of measurable sets, and hence $E$ is measurable.

**b) E has measure 0**

Since the measure of each $E_k$ is given by $m(E_k)=\frac{1}{k^2}$ we have $\sum_{k=1}^{\infty}m(E_k)\leq\sum_{k=1}^{\infty}\frac{1}{k^2}<\infty$. For any $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that $\sum_{k=N}^\infty m(E_k) < \varepsilon$. Now we have $$E=\bigcap_{n=1}^{\infty} \bigcup_{k=n}^\infty E_k\subseteq \bigcap_{n=N}^{\infty} \bigcup_{k=n}^\infty E_k\subseteq \bigcup_{k=N}^\infty E_k$$ and subadditivity gives $$m(E)=m(\bigcap_{n=1}^{\infty} \bigcup_{k=n}^\infty E_k) \leq m(\bigcap_{n=N}^{\infty} \bigcup_{k=n}^\infty E_k) \leq m(\bigcup_{k=N}^\infty E_k) \leq \sum_{k=N}^\infty m(E_k) < \varepsilon.$$ Hence $m(E)=0$.

### Problem 3 (Max):
**Let $f: \mathbb{R} \to \mathbb{R}$ be continuous. Show if $B$ is a Borel set, then $f(B)$ is, too.**

Let $f$ be a continuous function, and $I = (a,b)$ an open interval. Finally, let $I_n = [a+1/n, b-1/n]$. Note that we can write
$$
I = (a,b) = \bigcup_{n=1}^\infty [a+1/n, b-1/n]
$$
So, we can express the image of $I$ as
$$
f(I) = f\left(\bigcup_{n=1}^\infty I_n \right) = \bigcup_{n=1}^\infty f(I_n)
$$
Note that $I_n$ is compact, so its image under a continuous function is also compact, and in particular closed. So, $f(I)$ is a union of closed sets, which means that it is Borel. What's more, every open set $U \subseteq \mathbb{R}$ is a union of open intervals, so $f(U)$ is a union of Borel sets, making it Borel.

Now, let $S$ be the set of sets whose image under $f$ is Borel. Suppose we have some countable collection $\{S_i\}_{i\in\mathbb{N}} \subseteq S$. Note that $f\left( \bigcup_{i\in\mathbb{N}} S_i \right) = \bigcup_{i \in \mathbb{N}} f(S_i)$ and $f\left( \bigcap_{i\in\mathbb{N}} S_i \right) = \bigcap_{i \in \mathbb{N}} f(S_i)$, so the image of the union and intersection of these sets is a union or intersection of Borel sets, making them Borel. Thus, $S$ is closed under countable intersection and union.

Consider $A \in S$. If $f$ is injective, we have that $f(\mathbb{R}\setminus A) = f(\mathbb{R}) \setminus f(A)$. Thus, the image of $A^c = \mathbb{R} \setminus A$ is a difference of Borel sets, and thus Borel. Thus, $S$ is closed under taking complements. (how to show closure under complements without injectivity?). So, $S$ is a $\sigma$-algebra.

Finally, note that we have shown that $S$ contains all open sets. So, $S$ is a $\sigma$-algebra containing all of the open sets. Since the Borel sets form the smallest such $\sigma$-algebra, this means that all Borel sets are contained in $S$. In particular, if $B$ is Borel, then $f(B)$ is as well.

### Problem 4 (Austin):
**Let $f$ be an integrable function on $(0,1)$. Suppose that for any nonnegative continuous function $g: (0,1) \to \mathbb{R}$:
$$\int_0^1 fgdx \geq 0$$
Show that $f \geq 0$ almost everywhere.**

$Proof$

Suppose that for any nonnegative continuous function $g:(0,1)\rightarrow\mathbb{R}$
$$\int_0^1 fgdx \geq 0$$
Assume for contradiction that there exists set $E$ such that $m(E)\neq 0$ and $f(x)<0$ when $x\in E$. We will produce a nonnegative continuous function $g$ such that the inequality does not hold. Since $E$ is measureable, for every $\epsilon > 0$ there some open set, $\mathcal{O}$, such that $m(\mathcal{O}-E)<\epsilon$. We then have that $\mathcal{O}$ is the contable union of open intervals. So $\mathcal{O}=\bigcup_{i=0}^{n} \mathcal{O}_i$.

Define $g$ to be the piecewise linear function such that $g(x)=0$ when $x\notin\mathcal{O}$ and $g(x)=1$ when $x$ is the midpoint of some $\mathcal{O}_i$. We can now observe that $\int_{\mathcal{O}^c\cap(0,1)}fgdx=0$. So $$\int_0^1fgdx=\int_\mathcal{O}fgdx+\int_{\mathcal{O}^c\cap(0,1)}fgdx=\int_\mathcal{O}fgdx$$

Since $m(\mathcal{O}-E)<\epsilon$$$\int_\mathcal{O}fgdx=\int_E fgdx+\int_{\mathcal{O}-E}fgdx\leq\int_E fgdx+\epsilon c$$Since $\epsilon$ is arbitrary, we get $$\int_0^1fgdx=\int_\mathcal{O}fgdx=\int_E fgdx$$

Finally, since $\int_Efgdx<0$, we have that $\int_0^1 fgdx<0$ a contradiction. So $f\geq 0$ for $a.e.$ $x$

---

## June 2021 Prelim

### Problem 1 (Ian):
**Assume that $f:[0,1]\rightarrow\mathbb{R}$ is a continuous function. Define $f_n:[0,1]\rightarrow \mathbb{R}$ by $f_n(x) = f(x^n).$ Show that $f_n$ is integrable and compute $\displaystyle\lim_{n\rightarrow \infty}\int_{[0,1]}f_n(x)dx.$**

(Essential idea of the proof: Use the dominated convergence theorem, with the max of $|f(x)|$ acting as the upper bound for $f_n(x).$)

*Proof:*

First, note that for any fixed $n \in \mathbb{N}$, $0\leq x^n \leq x \leq 1$ and $f(x^n)$ is a continuous function on $[0,1]$. Additionally, since $f(x)$ is continuous on a compact set, it follows that $\displaystyle\max_{x \in [0,1]}|f(x)|$ exists and is finite. Moreover, we have that $$\displaystyle\max_{x\in[0,1]}|f_n(x)|=\displaystyle\max_{x\in[0,1]}|f(x)|,$$ and so $$\int_{[0,1]}|f_n(x)|dx = \int_{[0,1]}|f(x^n)|dx \leq \int_{[0,1]} \displaystyle\max_{x\in[0,1]}|f(x)| = \displaystyle\max_{x\in[0,1]}|f(x)|<\infty.$$ Therefore, for any given $n$, $f_n(x)$ is integrable on $[0,1]$. (Alternatively, you could also say that since $f_n(x)$ is continuous on a closed interval, it is Riemann integrable, and its Riemann integral equals its Lebesgue integral.)

The above also establishes that $|f_n(x)| \leq \displaystyle\max_{x\in [0,1]}|f(x)|$ for all $n$, noting that $\displaystyle\max_{x\in [0,1]}|f(x)|$ is integrable on $[0,1]$. Lastly, since $f(x)$ is continuous, we have for any fixed $x\in [0,1]$ that $$g(x):=\lim_{n\rightarrow \infty}f_n(x) =f\left( \lim_{n \rightarrow \infty}x^n\right) = \begin{cases}f(0), ~~~x\in [0,1)\\f(1),~~~x=1. \end{cases}$$ Consequently, we have by the Dominated Convergence Theorem that $$\lim_{n\rightarrow\infty}\int_{[0,1]}f_n(x)dx = \int_{[0,1]}\lim_{n\rightarrow \infty}f_n(x)dx = \int_{[0,1]}g(x)dx = f(0).$$

### Problem 2 (Ehren):
**Let $f: \mathbb{R}^d \to \mathbb{R}^d$ be a Lipschitz function (there is an $M > 0$ such that $|f(x)-f(y)| < M|x-y|$).**

**i) Find a constant $\tilde{M}$ such that for any cube $Q$, $m_*(f(Q)) < \tilde{M}|Q|$.**

**ii) Show that for any $E \subset \mathbb{R}^d$ with $m(E) = 0$, we have that $m(f(E)) = 0$**

*Proof:*

i) Let $Q \subset \mathbb{R}^d$ be a cube of side length $n$. Then the diameter of the cube is $n \cdot \sqrt{d}$. So the max distance between the image of two points is bounded above by $M\sqrt{d}$. Take $\tilde{M}$ to be $(2M\sqrt{d})^d$ if $M \geq 1$, and $1$ otherwise.

ii) Let $\varepsilon > 0$. Since the measure, and thus the outer measure, of $E$ is $0$, there is some covering of $E$ by almost disjoint closed cubes $\{Q_i\}_{i \in \mathcal{I}}$ such that $\sum_{i \in \mathcal{I}} |Q_i| < \frac{\varepsilon}{\tilde{M}}$.
Now using the first part, we know that:
$$
m(\bigcup_{i \in \mathcal{I}} f(Q_i))
< \sum_{i \in \mathcal{I}} \tilde{M}|Q_i|
= \tilde{M} \sum_{i \in \mathcal{I}} |Q_i|
< \tilde{M} \cdot \frac{\varepsilon}{\tilde{M}}
= \varepsilon
$$

### Problem 3 (Emma & Erin):
**Let $f(x,y)$ be a nonnegative and measurable function in $\mathbb{R}^2$. Suppose that for a.e. $x\in \mathbb{R}, f(x,y)$ is finite for a.e. $y \in \mathbb{R}$. Show that for a.e. $y \in \mathbb{R}, f(x,y)$ is finite for a.e. $x\in \mathbb{R}$.**

*Proof:* Let $E=\{(x,y)\in\mathbb{R}^2\mid f(x,y)=\infty\}$. Because $f$ is measurable we know that the set of points where $f$ is finite is measurable, and therefore the complement, namely $E$, is a measurable set in $\mathbb{R}^2$. Then the indicator function $\mathbb{1}_E(x,y)$ is a nonnegative, measurable function on $\mathbb{R}^2$, so by Tonelli's Theorem, we have $$\int_{\mathbb{R}^2}\mathbb{1}_E(x,y)\,dx\,dy=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\mathbb{1}_E(x,y)\,dx\right)\,dy=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\mathbb{1}_E(x,y)\,dy\right)\,dx.$$ For almost every $x\in\mathbb{R}$ we have that the set $\{y\in\mathbb{R}\mid f(x,y)=\infty\}$ has measure $0$. Then $\int_{\mathbb{R}}\mathbb{1}_E(x,y)\,dy=0$ for almost every $x$, so we have $$0=\int_{\mathbb{R}}0\,dx=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\mathbb{1}_E(x,y)\,dy\right)\,dx=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\mathbb{1}_E(x,y)\,dx\right)\,dy.$$ Then we have $0=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\mathbb{1}_E(x,y)\,dx\right)\,dy,$ which implies that $0=\int_{\mathbb{R}}\mathbb{1}_E(x,y)\,dx=m\left(\{x\in\mathbb{R}\mid f(x,y)=\infty\}\right)$ for almost every $y\in\mathbb{R}$ by Property $5$ of Lebesgue integrals. We conclude that for a.e. $y\in\mathbb{R}$ we have $f(x,y)<\infty$ for a.e. $x\in\mathbb{R}$.

### Problem 4 (Evan):
**Let $\{f_k\}$ be a sequence of nonnegative measurable functions on $\mathbb{R}$. Suppose $f_k \to f$ and $f_k \leq f$ almost everywhere. Show the following:
$$\int_\mathbb{R} f_k dx \to \int_\mathbb{R} f dx$$**

*Proof:*

Since $f_k \leq f$ almost everywhere it follows that
$$\int_{\mathbb{R}}f_k dx\leq \int_{\mathbb{R}} f dx$$
for all $k$. Hence,
$$\limsup_{k \to \infty} \int_{\mathbb{R}}f_k dx\leq \int_{\mathbb{R}} f dx.$$
Since we are given that $f_k \geq 0$ and $f_k \to f$ almost everywhere we can apply Fatous Lemma to conclude,
$$\int_{\mathbb{R}} f dx \leq \liminf_{k \to \infty} \int_{\mathbb{R}}f_k dx.$$
Combining the 2 previoius inequalites we obtain,
$$\int_{\mathbb{R}} f dx \leq \liminf_{k \to \infty} \int_{\mathbb{R}}f_k dx \leq
\limsup_{k \to \infty} \int_{\mathbb{R}}f_k dx\leq \int_{\mathbb{R}} f dx.
$$
Hence we have equality all the way through giving that,
$$
\int_{\mathbb{R}} f dx
=\liminf_{k \to \infty} \int_{\mathbb{R}}f_k dx
= \limsup_{k \to \infty} \int_{\mathbb{R}}f_k dx
= \lim_{k \to \infty} \int_{\mathbb{R}}f_k dx
$$
Therefore we have shown that $\int_{\mathbb{R}}f_k dx \to \int_{\mathbb{R}} f dx$

---

## June 2022 Prelim

### Problem 1 (Evan):
**Let $f,g$ be two integrable functions such that for every measurable set $E$,
$$\int_E f \leq \int_E g$$
Show $f \leq g$ almost everywhere**

*Proof:*

Note that for any ball, $B \in \mathbb{R}^d$ we have that $B$ is measurable therefore $$\int_B f(x)dx \leq \int_B g(x)dx.$$
Moreover, $$\frac{1}{m(B)}\int_B f(x)dx \leq \frac{1}{m(B)} \int_B g(x)dx.$$
Since this is true for all balls in $\mathbb{R}^d$ we have that for any $x \in \mathbb{R}^d$ we can consider balls containing $x$ as for any $x \in \mathbb{R}^d$ we have the following inequality.
$$\lim_{\substack{m(B) \to 0 \\ x \in B}}\frac{1}{m(B)}\int_B f(y)dy \leq \lim_{\substack{m(B) \to 0 \\ x \in B}}\frac{1}{m(B)} \int_B g(y)dy.$$
Recall that $f,g \in L^1(\mathbb{R}^d)$ therefore the Lebesgue Differentiation Lemma gives,
$$\lim_{\substack{m(B) \to 0 \\ x \in B}}\frac{1}{m(B)}\int_B f(y)dy = f(x)
\qquad \text{and} \qquad
\lim_{\substack{m(B) \to 0 \\ x \in B}}\frac{1}{m(B)} \int_B g(y)dy = g(x),
$$
for almost every $x \in \mathbb{R}^d$. Therefore the inequality above gives us,
$$f \leq g$$
almost everywhere.

### Problem 2 (Luke):
**Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuously differentiable function whose derivative is bounded on $\mathbb{R}$, and $g : \mathbb{R} \rightarrow \mathbb{R}$ is Lebesgue integrable. Show that
$$ (f * g)(x) = \int_{\mathbb{R}} f(x - y)g(y) \,dy $$
is a differentiable function on $\mathbb{R}$, and
$$ (f * g)'(x) = \int_{\mathbb{R}} f'(x - y)g(y) \,dy. $$**

*(Proof Idea: Use the Mean-Value Theorem to bound each difference quotient by an integrable function. Then, use the Dominated Convergence Theorem to pass a limit through the integral in order to get the difference quotient to tend towards the derivative.)*

*Proof.* Because $f'$ is bounded on $\mathbb{R}$, there must exist some $M > 0$ such that $|f'(x)| \leq M$ for all $x \in \mathbb{R}$. For any fixed $x \in \mathbb{R}$ and $h > 0$, we can use the Mean-Value Theorem to guarentee the existence of some $c \in (x, x + h)$ such that
$$ \left|\frac{f(x + h) - f(x)}{h}\right| = |f'(c)| \leq M. $$
Similarly, for any fixed $x \in \mathbb{R}$ and $h < 0$, we can find some $c \in (x + h, x)$ such that
$$ \left|\frac{f(x) - f(x + h)}{h}\right| = |f'(c)| \leq M. $$

Thus, for any fixed $x \in \mathbb{R}$ and $h \in \mathbb{R} \setminus \{ 0 \}$, we have that for all $y \in \mathbb{R}$,
$$ \left|\frac{f(x - y + h) - f(x - y)}{h} g(y) \right| = \left|\frac{f(x - y + h) - f(x - y)}{h}\right||g(y)| \leq M|g(y)|. $$
And because $\int_{\mathbb{R}} |Mg(y)| \, dy = M \int_{\mathbb{R}} |g(y)| \, dy < \infty$, we see that each difference quiotient is bounded above by the same integrable function. So we can apply the Dominated Convergence Theorem to pass a limit through an integral. That is, for any $x \in \mathbb{R}$,
\begin{align*}
\lim_{h \rightarrow 0} \frac{(f * g)(x + h) - (f * g)(x)}{h} &= \lim_{h \rightarrow 0} \frac{1}{h} \left( \int_{\mathbb{R}} f(x + h - y)g(y) \, dy - \int_{\mathbb{R}} f(x - y)g(y) \, dy\right) \\
&= \lim_{h \rightarrow 0} \int_{\mathbb{R}} \frac{f(x - y + h) - f(x - y)}{h}g(y) \, dy \\
&= \int_{\mathbb{R}} \lim_{h \rightarrow 0}\frac{f(x - y + h) - f(x - y)}{h}g(y) \, dy \\
&= \int_{\mathbb{R}} f'(x - y)g(y) \, dy.
\end{align*}
Because the limit as $h$ tends towards $0$ of the difference quiotient exists, we can say that $(f * g)$ is a differentiable function on $\mathbb{R}$, and $(f * g)'(x) = \int_{\mathbb{R}} f'(x - y)g(y) \,dy$. $\;\;\;\square$

### Problem 3 (Ian):

**Suppose that $f:\mathbb{R}\rightarrow \mathbb{R}$ is integrable on $\mathbb{R}$. Show that $$\lim_{n\rightarrow \infty} \int_{\mathbb{R}}f(x)\sin(nx)dx=0.$$ Hint: You may use the fact that step functions are dense in $L^1(\mathbb{R}).$**

*Proof:*

(Essential idea of the proof: Pick a step function that approximates $f(x)$ in the $L^1$ norm. Show that when this step function is integrated against $\sin(nx)$, the integral vanishes for large $n$. To show the desired limit, add an intelligent 0 to the integral and then use triangle inequality.)

Let $\epsilon>0$ be given. Since step functions are dense in $L^1(\mathbb{R}),$ there exists a $$\chi(x) = \sum_{i=1}^N a_i\chi_{I_{i}}(x)$$ such that $\|f-\chi\|_{L^1(\mathbb{R})}<\frac{\epsilon}{2}$, where we may assume without loss of generality that the intervals $I_i = [b_i,c_i]$ are almost disjoint with the property that for all $x \in I_i$ and $y \in I_j$, if $i<j$, then $x \leq y$ (I am arranging the intervals "in order from smallest elements to largest elements"). Then note that for any $n$,
\begin{align*}
\int_{\mathbb{R}}\chi(x)\sin(nx)dx &= \int_{\mathbb{R}}\left(\sum_{i=1}^N a_i\chi_{I_i}(x)\right) \sin(nx) = \sum_{i=1}^N a_i \int_{\mathbb{R}}\chi_{I_i}(x)\sin(nx)dx\\ &=\sum_{i=1}^N a_i \int_{[b_i,c_i]}\sin(nx)dx = \sum_{i=1}^N \frac{a_i}{n}(\cos(b_i)-\cos(c_i)).
\end{align*}
Then we have that $$\left|\int_{\mathbb{R}}\chi(x) \sin(nx)dx\right| \leq \frac{1}{n} \sum_{i=1}^N 2|a_i|\leq \frac{1}{n}\sum_{i=1}^N 2\max_{k}(|a_k|),$$ and so there exists an $N' \in \mathbb{N}$ sufficiently large such that for all $n \geq N'$, $$\left|\int_{\mathbb{R}}\chi(x) \sin(nx)dx\right|<\frac{\epsilon}{2}.$$ Therefore, for all $n \geq N'$, it follows that
\begin{align*}
\left|\int_{\mathbb{R}}f(x)\sin(nx)dx\right| &=\left| \int_{\mathbb{R}}(f(x)-\chi(x))\sin(nx)dx + \int_{\mathbb{R}}\chi(x)\sin(nx)dx \right|\\
&\leq \int_{\mathbb{R}}|f(x)-\chi(x)| |\sin(nx)|dx +\left| \int_{\mathbb{R}}\chi(x)\sin(nx)dx \right| \\
&\leq \int_{\mathbb{R}}|f(x)-\chi(x)|dx +\left| \int_{\mathbb{R}}\chi(x)\sin(nx)dx \right|\\
&<\frac{\epsilon}{2}+\frac{\epsilon}{2} =\epsilon.
\end{align*}

We then conclude that $$\lim_{n\rightarrow \infty} \int_{\mathbb{R}}f(x)\sin(nx)dx=0.$$

### Problem 4 (Pablo):
**Let $\{f_k\}$ be a sequence of measurable functions defined on a measurable set $E \subset \mathbb{R}^d$ with $m(E) < \infty$.
Suppose that for each $x \in E$,
$$ \sup \{ |f_k(x)| : k \geq 1 \} < \infty. $$
Show that for each $\epsilon > 0$ there exists a closed set $F$ such that $m(E \setminus F) < \epsilon$ and
$$ \sup \{ |f_k(x)| : x \in F \text{ and } k \geq 1 \} < \infty. $$**

*Proof.*
Let $f(x) = \sup \{|f_k(x)| : k \geq 1\}$ for all $x \in E$.
Since $\{f_k\}$ is a sequence of measurable functions, $f$ is also a measurable function.
Furthermore, $\sup \{|f_k(x)| : k \geq 1\} < \infty$ for each $x \in E$, so $f$ is finite-valued.

Let $\epsilon > 0$ be given.
For each $n \in \mathbb{N}$, let $E_n := \{x : f(x) \leq n\}$.
Since $f$ is a measurable function, each $E_n$ is a measurable set.
Since $f$ is finite-valued, $\lim_{n \to \infty} E_n \uparrow E$.
Thus $\lim_{n \to \infty} m(E_n) \uparrow m(E)$, and $\lim_{n \to \infty} m(E \setminus E_n) \downarrow 0$ since $m(E) < \infty$.
Therefore there exists an $M \in \mathbb{N}$ such that $m(E \setminus E_M) < \frac{\epsilon}{2}$.

Since $E_M$ is measurable, there exists a closed set $F \subseteq E_M$ such that $m(E_M \setminus F) < \frac{\epsilon}{2}$.
Also, since $F \subseteq E_M \subseteq E$ we have $E \setminus F = (E \setminus E_M) \cup (E_M \setminus F)$.
Therefore,$$
m(E \setminus F)
= m((E \setminus E_M) \cup (E_M \setminus F))
\leq m(E \setminus E_M) + m(E_M \setminus F)
< \frac{\epsilon}{2} + \frac{\epsilon}{2}
= \epsilon
$$
and $|f_k(x)| \leq |f(x)| \leq M < \infty$ for all $x \in F$ and $k \geq 1$.