Show that if $f:\mathbb{R}^2\rightarrow\mathbb{R}$ is measurable (with respect to the Lebesgue-measurable sets in $\mathbb{R}^2)$, then there exists a Borel function $g$ such that $f(x)=g(x)$ for almost every $x\in\mathbb{R}^2$ (i.e. for all $x\in\mathbb{R}^2$ except a set of measure zero).

I guess "Borel function" means Borel-measurable functions.

If so, $g$ Borel-measurable would mean that $g^{-1}(A)$ is a Borel set in $\mathbb{R}^2$ for all Borel $A\in\mathbb{R}$. And $f$ Lebesgue-measurable would mean that $f^{-1}(A)$ is a Lebesgue-measurable set in $\mathbb{R}^2$ for all Borel $A\in\mathbb{R}$.

Given this setting, it is hard to see how to proceed. I am allowed to change the value of $f(x)$ for a subset of measure zero in $\mathbb{R}^2$, and I want to get a Borel-measurable function. How can I do that?

  • $\begingroup$ Your guess is right. Do you know that every Lebesgue-measurable set is the union of a Borel set and a Lebesgue null set? $\endgroup$ – Daniel Fischer Oct 11 '13 at 21:45
  • $\begingroup$ Daniel has a nice suggestion here: math.stackexchange.com/a/459853/27978 $\endgroup$ – copper.hat Oct 11 '13 at 22:29

Every Lebesgue-measurable set is the union of a Borel set (one can choose an $F_\sigma$ for that) and a Lebesgue null set.

The pointwise limit of a sequence of Borel-measurable functions is Borel-measurable.

Combining the two leads to the result. For $n \in \mathbb{N}$ and $k \in \mathbb{Z}$, let

$$E_{n,k} = f^{-1}\left(\left[\frac{k}{2^n},\frac{k+1}{2^n} \right)\right).$$

Decompose $E_{n,k}$ into a Borel set $B_{n,k}$ and a null set $N_{n,k}$. Define $g_n$ as $\frac{k}{2^n}$ on $B_{n,k}$, and as $0$ on $N_{n,k}$. Show that the so-defined $g_n$ is Borel measurable. Show that $g_n \to f$ almost everywhere.

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