This post is a short note regarding problem 14 from Page-59 of Rudin's Real and Complex Analysis.
As stated the proposition is incorrect. Best we can do is find Borel functions $g$ and $h$ such that $g(x) = h(x)$ a.e. $[\mu_k]$ and $g(x) \le f(x) \le h(x)$ a.e. $[\mu_k]$.
To see why it's not always possible to 'approximate' a lebesgue measurable function from both above and/or below by Borel functions, we rely on a cardinality argument. Roughly speaking, there aren't enough Borel functions to account for every modification of a lebesgue measurable function in a measure zero set.
Lemma: The cardinality of real Borel measurable functions on $\mathbb{R}$ is $2^{\aleph_0}$.
Proof: Since, constant functions are Borel measurable there are atleast $2^{\aleph_0}$ Borel functions. Again, to determine a Borel function $g$ it suffices to determine the sequence of sets $g^{-1}(q,\infty)$ for each $q \in \mathbb{Q}$ (ordered by inclusion). Since, there are $2^{\aleph_0}$ (i.e., the cardinality of Borel subsets of $\mathbb{R}$)-choices for each $g^{-1}(q,\infty)$ as $q$ varies over $\mathbb{Q}$, there are atmost $\left(2^{\aleph_0}\right)^{\aleph_0} = 2^{\aleph_0}$ real Borel measurable functions on $\mathbb{R}$. $\boxed{}$
Let us index the Borel functions by $\{g_j: j \in J\}$, where, $J$ is some indexing set with cardinality $2^{\aleph_0}$.
Start with the Cantor set $C \subset \mathbb{R}$ which is a set of lebesgue measure zero. Let $C = C_1 \sqcup C_2$ be a partition into subsets such that $|C_1| = |C_2| = 2^{\aleph_0}$. Let us index the elements of $C_1 = \{x_j : j \in J\}$ and $C_2 = \{x_j' : j \in J\}$. Now, if we consider a functions $f: C \to \mathbb{Z}$ such that $f(x_j) < g_j(x_j)$ and $f(x_j') > g_j(x_j')$ for each $j \in J$ (such a $f$ is possible to construct since $\mathbb{Z}$ is an unbounded discrete subset of $\mathbb{R}$), and extend it to $\mathbb{R}$ by defining $f(x) = 0$ for $x \in \mathbb{R}\setminus C$, we have constructed a lebesgue measurable function $f$ which violates the requirement of the proposition form the book.
N.B.: If we were given the scenario that $f$ is a 'bounded' real-lebesgue measurable function on $\mathbb{R}$ (or if, $f:\mathbb{R} \to [-\infty,\infty]$ is lebesgue measurable on the extended real-line), then it is possible to construct borel functions (respectively, extended Borel functions) with the required property that is $g \le f \le h$ for all $x \in \mathbb{R}$. As long as range of $f$ has unbounded discrete subset we may modify $f$ in the above fashion inside an uncountable subset of measure zero such that it is no longer possible to be bounded by Borel functions at all points.
Problem: Let, $f$ be a real-valued Lebesgue measurable function on $\mathbb{R}^k$, prove that there exists Borel functions $g$ and $h$ such that $\mu_k(\{x \in \mathbb{R}^k: g(x) \neq h(x)\}) = 0$ and $g(x) \le f(x) \le h(x)$ for all $x \in \mathbb{R}^k$. (where, $\mu_k$ is the Lebesgue measure on $\mathbb{R}^k$)
As stated the proposition is incorrect. Best we can do is find Borel functions $g$ and $h$ such that $g(x) = h(x)$ a.e. $[\mu_k]$ and $g(x) \le f(x) \le h(x)$ a.e. $[\mu_k]$.
To see why it's not always possible to 'approximate' a lebesgue measurable function from both above and/or below by Borel functions, we rely on a cardinality argument. Roughly speaking, there aren't enough Borel functions to account for every modification of a lebesgue measurable function in a measure zero set.
Lemma: The cardinality of real Borel measurable functions on $\mathbb{R}$ is $2^{\aleph_0}$.
Proof: Since, constant functions are Borel measurable there are atleast $2^{\aleph_0}$ Borel functions. Again, to determine a Borel function $g$ it suffices to determine the sequence of sets $g^{-1}(q,\infty)$ for each $q \in \mathbb{Q}$ (ordered by inclusion). Since, there are $2^{\aleph_0}$ (i.e., the cardinality of Borel subsets of $\mathbb{R}$)-choices for each $g^{-1}(q,\infty)$ as $q$ varies over $\mathbb{Q}$, there are atmost $\left(2^{\aleph_0}\right)^{\aleph_0} = 2^{\aleph_0}$ real Borel measurable functions on $\mathbb{R}$. $\boxed{}$
Let us index the Borel functions by $\{g_j: j \in J\}$, where, $J$ is some indexing set with cardinality $2^{\aleph_0}$.
Start with the Cantor set $C \subset \mathbb{R}$ which is a set of lebesgue measure zero. Let $C = C_1 \sqcup C_2$ be a partition into subsets such that $|C_1| = |C_2| = 2^{\aleph_0}$. Let us index the elements of $C_1 = \{x_j : j \in J\}$ and $C_2 = \{x_j' : j \in J\}$. Now, if we consider a functions $f: C \to \mathbb{Z}$ such that $f(x_j) < g_j(x_j)$ and $f(x_j') > g_j(x_j')$ for each $j \in J$ (such a $f$ is possible to construct since $\mathbb{Z}$ is an unbounded discrete subset of $\mathbb{R}$), and extend it to $\mathbb{R}$ by defining $f(x) = 0$ for $x \in \mathbb{R}\setminus C$, we have constructed a lebesgue measurable function $f$ which violates the requirement of the proposition form the book.
N.B.: If we were given the scenario that $f$ is a 'bounded' real-lebesgue measurable function on $\mathbb{R}$ (or if, $f:\mathbb{R} \to [-\infty,\infty]$ is lebesgue measurable on the extended real-line), then it is possible to construct borel functions (respectively, extended Borel functions) with the required property that is $g \le f \le h$ for all $x \in \mathbb{R}$. As long as range of $f$ has unbounded discrete subset we may modify $f$ in the above fashion inside an uncountable subset of measure zero such that it is no longer possible to be bounded by Borel functions at all points.
