Following is exercise 2.14 in Rudin's RCA.
Let $f$ be a real-valued Lebesgue measurable function on $\mathbb{R}^{k}$. Prove that there exist Borel functions $g$ and $h$ such that $g(x)=h(x)$ a.e. $\mbox{$[m]$}$, and $g(x) \le f(x) \le h(x)$ for every $ x \in \mathbb{R}^{k}$. (Here, $m$ denotes the Lebesgue measure on $\mathbb{R}^{k}$.)
What I tried is written below.
If we think $-f$ instead of $f$, it is enough to find Borel function $g$ such that $g(x)=f(x)$ a.e. $\mbox{$[m]$}$, and $g(x) \le f(x)$. If $f \in L^{1}(\mathbb{R}^{k})$, by Vitali-Carathéodory theorem, there exists a sequence of upper semicontinuous functions $\{u_n\}$ such that $u_n \le f$ and $$\int_{\mathbb{R}^{k}}(f-u_n )dm < \frac{1}{n}$$ Define $g=\sup_{n} u_n$ It is trivial that $g \le f$ and $f=g$ a.e. Only thing we need is that $g$ is Borel function. We have $$ \{x: u_n (x) < \alpha \}, \\ \{x: u_n (x) \le \alpha \} = \bigcap_{m=1}^{\infty} \{x: u_n (x) < \alpha + \frac{1}{m} \}, \\ \{x: g(x) \le \alpha \} = \bigcap_{n=1}^{\infty} \{x: u_n (x) \le \alpha \}, \\ \{x: g(x) < \alpha \} = \bigcup_{m=1}^{\infty} \{x: g(x) \le \alpha - \frac{1}{m} \}, \\ \{x: g(x) > \alpha \} = {\{x: g(x) \le \alpha \}}^{\mathsf{c}} $$ are Borel sets. Thus $g$ is Borel function, and this concludes the case when $f \in L^{1}(\mathbb{R}^{k})$. Now split $\mathbb{R}^{k}$ to countable number of cubes with side length $1$. Call them $\{B_{n}\}$. Since $B_n \cap f^{-1}([m, m+1))$ is Lebesgue measurable set, there exists $B_{n, m} \subset B_n \cap f^{-1}([m, m+1))$ such that $B_{n, m}$ is a $F_{\sigma}$ set, thus a Borel set, and $m(B_{n, m}) = m(B_n \cap f^{-1}([m, m+1)))$. Split each $B_n$ into $\{B_{n, m}\}_{m \in \mathbb{Z}}$ and $B_n - \bigcup_{m \in \mathbb{Z}} B_{n, m}$ By definition of $B_{n, m}$, $f$ is summable on $B_{n, m}$, and because $$ m(B_n)=\sum_{m \in \mathbb{Z}} m(B_n \cap f^{-1}([m, m+1)) )=\sum_{m \in \mathbb{Z}}m(B_{n, m}), \\ m(B_n - \bigcup_{m \in \mathbb{Z}} B_{n, m})=0, $$ $f$ is also summable on $B_n - \cup B_{n, m}$. Thus for each part($B_{n, m}$ or $B_n - \cup B_{n, m}$), we can choose $g$ by using previous discussion. Since these sets are all Borel sets, gluing these functions $g$ again gives a Borel function. $g \le f$ and $f=g$ a.e. are obvious, since we splitted $\mathbb{R}^{k}$ into countable number of parts.
Is my proof valid? I'm not sure because it feels like I used Vitali-Carathéodory for non-summable function. (Of course we cannot prove Vitali-Carathéodory thm for non-summable function by using above idea, because we cannot choose OPEN sets inside Lebesgue measurable set which has same measure.)
Any kind of comments will be appreciated.