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.
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