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Suppose $f$ is a given Lebesgue measurable function on $\Bbb{R}$. Show that there exists Borel measurable functions $g$ and $h$ satisfying $g=h$ a.e. and $g(x)\le f(x) \le h(x)$ for every $x \in \Bbb{R}$

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Hint: enumerate the rationals. For each rational $r$, there are Borel sets $L(r)$ and $U(r)$ such that $L(r) \subseteq \{x: f(x) \le r\} \subseteq U(r)$ and $m(U(r) \backslash L(r)) = 0$. Using the enumeration, we can do this so that $L(s) \subseteq L(r)$ and $U(r) \subseteq U(s)$ if $s < r$. Construct your $g$ and $h$ using these.

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    How do I construct $L(r)$ and $U(r)$ such that $L(s) \subseteq L(r)$ and $U(r) \subseteq U(s)$ for $s – 2017-02-15
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    By induction. There are finitely many previous rationals in the enumeration, which require adjusting on sets of measure $0$. – 2017-02-15
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    Okay, I got that point. And sir for the end part, will putting $L(r)=g^{-1}((-\infty,r))$ and $U(r)=h^{-1}((-\infty,r))$ work? I'm not very sure – 2017-02-16