Measure Theory – Does Lebesgue Measurable Area Imply Function Measurability?

Question

The opposite implication is true, but i don't know about this side. I've considered intersecting the area under the graph of the function with subsets of the form $\mathbb{R} \times (a,\infty)$. This intersection has to be measurable, but it doesn't seem to follow that the x-coordinates of this set are also measurable (in $\mathbb{R}$). For example, the set $\mathcal{V} \times \{ 0 \}$, where $\mathcal{V}$ is a vitali set, is measurable on $\mathbb{R}^2$, but it's x-coordinates are not measurable on $\mathbb{R}$.

Answer 1

This is a little too long for a comment.

When considering in $\Bbb R^2$ the product of the Lebesgue $\sigma$-algebra for $\Bbb R$, the question is already answered by @Kavi Rama Murthy here: If the area under graph of $f$ is measurable then $f$ is measurable .

So the step that remains is due to the fact that the Lebesgue $\sigma$-algebra for $\Bbb R^2$ is NOT the product of the Lebesgue $\sigma$-algebra for $\Bbb R$. It is the completion of the product $\sigma$-algebra.

Let $f: \Bbb R \rightarrow [0,\infty]$ be a function. Let $E=\{(x,y)\in \Bbb R^2:0 < y < f(x)\} $ and suppose that $E$ is Lebesgue measurable in $\Bbb R^2$.

For any $a \in [0, \infty]$, let us define the section $E^{a} = \{x:(x,a)\in E\}$. It is easy to see that $E^{a} = \{x:(x,a)\in E\}= \{x: f(x) >a\}$.

By Fubini Theorem for completion of product $\sigma$-algebras, we have that for almost all $a\in [0, \infty]$, $E^a$ is measurable. That means, there exist a set $N \subset [0, \infty]$, such that $\lambda(N) =0$ and for all $a \in [0, \infty] \setminus N$, $E^a$ is measurable.

So it remains to prove that for all $a\in N$, we also have that $E^a$ is measurable.

Note that $E^{\infty} = \emptyset$ and it is measurable. So, we can assume, without loss of generality that $N \subset [0, \infty)$,

Given $a \in N$. Then $0 \leq a < \infty$. Since $\lambda(N) =0$, given any $\varepsilon >0$, we have that $(a, a +\varepsilon) \nsubseteq N$. So, we can build a strictly decreasing sequence $(a_n)_n$ such that for all $n$, $a_n \in (a, \infty) \setminus N$ and $a_n \to a$.

So, for all $n$, $E^{a_n}$ is measurable and $$E^{a} = \{x: f(x) >a\}= \bigcup_n \{x: f(x) >a_n\} =\bigcup_n E^{a_n}$$ So $E^a$ is measurable.

So we have proved that for all $a \in [0, \infty]$ , $E^a$ is measurable. Since $E^{a} =\{x: f(x) >a\}$, we have that for all $a \in [0, \infty]$, $\{x: f(x) >a\}$ is measurable, which means that $f$ is Lebesgue measurable.