[Pugh] Chapter 6 Exercise 31: Prove that a measurable function $f$ is sandwiched as $u\leq f\leq v$, where $u$ is upper semicontinuous, $v$ is lower semicontinuous (we permit $v(x) = \infty$ at some points), and $v - u$ has small integral.
By exercise 30, we know that given a compact set $K \subseteq \mathbb{R} \times [0, \infty)$, then we have a upper semicontinuous $$u(x)=\begin{cases} \max\{y : (x,y) \in K\} & \text{if } K \cap (x \times \mathbb{R}) \neq \emptyset, \\ 0 & \text{otherwise}. \end{cases}$$
Proof: Suppose $0\le f\le M$ and is defined on $[a,b]$. Since the undergraph of $f$ defined as $\mathcal{U}f=\{(x,y):x\in[a,b],0\le y\le f(x)\}$, is measurable, there exists a closed set $K_1$ such that $K_1\subset \mathcal{U}f$ and the measure $m(\mathcal{U}f\setminus K_1)<\frac{\epsilon}{2}$. Let $$u(x)=\begin{cases} \max\{y : (x,y) \in K_1\} & \text{if } K_1 \cap (x \times \mathbb{R}) \neq \emptyset, \\ 0 & \text{otherwise}. \end{cases}$$ Similarly, let $g=M-f$, then $g$ is measurable. Since $\mathcal{U}g$ is measurable, there exists a closed set $K_2$ such that $K_2\subset \mathcal{U}g$ and the measure $m(\mathcal{U}g\setminus K_2)<\frac{\epsilon}{2}$. Let $$\tilde{v}(x)=\begin{cases} \max\{y : (x,y) \in K_2\} & \text{if } K_2 \cap (x \times \mathbb{R}) \neq \emptyset, \\ 0 & \text{otherwise}. \end{cases}$$ Then, $u\le f$ and $\tilde{v}\le g$ are upper semicontinuous. Let $v=M-\tilde{v}\ge f$, which is lower semicontinuous. We have $$\int(v-u)=m(\mathcal{U}f\setminus K_1)+m(\mathcal{U}g\setminus K_2)<\epsilon.$$
I was told to approach the problem as shown above, but still curious about not all Lebesgue measurable functions are bounded almost everywhere, then why can I suppose $0\le f\le M$? Also, the proof does not utilize the assumption that $v(x) = \infty$ at some points. What is the gap in the proof? Or, is there any property that I've misunderstood?