I’ve been slacking off writing my solutions here. Here are four solutions from Chapter 2. Hugo new blog/blog-post.md $\textbf{Problem 3.}$ If $\{f_n\}$ is a sequence of measurable functions on $X$, then $\left\{x: \lim f_n(x)\right.$ exists $\}$ is a measurable set.

$\textbf{Proof.}$ Since $\{f_n\}$ is a measurable sequence, so are the functions $g(x):=\limsup f_n(x)$ and $h(x):=\liminf f_n(x)$. Define the following new function $l(x)=g(x)-h(x)$. If $l(x)$ when it is not defined, let it equal $a \in \mathbb{\overline{R}}$.

$l(x)$ is a measurable function. $\{x: l(x) = 0\}$ is measurable. This is also exactly the set where $\{x: \lim f_n(x)\}$ exists.

$\textbf {Problem 4}$ if $ f: X \rightarrow \overline{\mathbb{R}} \text { and } f^{-1}((r, \infty]) \in \mathcal{M} \text { for each } r \in \mathbb{Q} \text {, then } f \text { is measurable. }$

$\textbf{Solution.}$ Let $\mathcal{G}=f^{-1}((r,\infty])$. It is obvious that $\mathcal{G} \subset \mathcal{M}$ so that $\sigma(G) \subset \mathcal{M}$.

It is also obvious that $f^{-1}((a,\infty]) \subset \mathcal{G}$ because the rationals are dense in the reals (think countable operations). I.e., if we let $r_n \rightarrow a$, we have that

$$\cup_{n=1}f_n^{-1} ((r_n,\infty]) = f^{-1}((a,\infty])$$.

$\textbf{Problem 6.}$ The supremum of an uncountable family of measurable $\overline{\mathbb{R}}$-valued functions on $X$ can fail to be measurable (unless the $\sigma$-algebra $\mathcal{M}$ is very special).

$\textbf{Solution.}$ Let’s consider the the Borel space $(\mathbb{R}, \mathbb{B}_{\mathbb{R}})$. We know there is a set $L$ that is not Borel. Let’s define the following functions $f_x: \mathbb{R}\rightarrow \overline{\mathbb{R}}$ so that
$$
f_x(y) =
\begin{cases}
1 & \text{if } y=x \

0 & \text{otherwise}
\end{cases}
$$
for all $x \in \mathbb{R}$. These functions are all Borel measurable. However, consider the following function

$$ g(y)=\sup_{x \in L}f_x(y) = \begin{cases}
1 & \text{if } y \in L \

0 & \text{otherwise}
\end{cases}
$$

which isn’t measurable.

$\textbf{Problem 8. } f: \mathbb{R} \rightarrow \mathbb{R} \text { is monotone, then } f \text { is Borel measurable. }$

$\textbf{Solution.}$ Let’s assume WLOG that $f$ is monotonically increasing. Consider all $x$ s.t. $f(x)<\alpha$. I claim that it is an open set in $\mathbb{R}$. Let’s denote that set by $A$. If it is empty, we are done. Assume it is therefore not empty. Let $y \in A$. For any neighborhood to the left of $y$, we’ll have that $f(x)<f(y)<\alpha$. The question is whether we can find a neighborhood to the right of $y$.

Notice that $\sup A \le \alpha$ and that there isn’t an $x \in A$ such that $f(x) = \alpha$. Therefore, we can always find $\{y_n\}$ in $A$ such that $y_n \rightarrow \sup A$ and $y_n \in A$. It is obvious that there is an $n$ such that $y<y_n$. We can define $r:=y_n-y$ to be our neighborhood so that $N_r(y) \subset A$.