I was thinking if this simple argument can be used to show that there are lebesgue measurable sets, that are not borel measurable.

Let $E \subset \mathbb{R}^{N+M}$ be a borel measurable set, then we can easily prove that the slices $E_x = \{ y \in \mathbb{R}^M : (x,y) \in E\}$ are borel measurable for each $x \in R^N$. This can be done by seeing that the claim is true for every box in $\mathbb{R}^{N+M}$, and proving that $M = \{ E \in \mathbb{R}^{N+M} : E_x\ \text{is borel measurable for all}\ x\in \mathbb{R}^N \}$ is a $\sigma$-algebra.

Now assume that all Lebesgue measurable sets in $\mathbb{R}^{N+M}$ are borel measurable, thus we get that all the slices $E_x$ of a Lebesgue measurable sets, are borel, and thus Lebesgue, measurable. This is a contradiction, in fact, for $N \ge 1$ and $M \ge 1$ we can generalize the following construction:

Take $V$ the Vitali set in $\mathbb{R}$ consider $\{0\}\ \text{x}\ V= E$. It's easy to see that $E \subset \mathbb{R}^2$ is Lebesgue measurable, being a set of outer measure zero, but the slice $E_0$ is the Vitali set in $\mathbb{R}$ and thus not lebesgue (borel) measurable.

I think it should work as a proof that there are, in $\mathbb{R}^N$ for $N \ge 1$, lebesgue measurable sets, wich are not borel measurable.

What do you think?

  • $\begingroup$ Your (correct) proof ensures the existence of a Lebesgue measurable set which is not Borel measurable inside $\Bbb{R}^2$: but what about inside $\Bbb{R}$? $\endgroup$ – Crostul Jan 7 '17 at 15:15
  • $\begingroup$ I know, the proof obviously doesn't work in $\mathbb{R}$, but this constuction is nice, according to me, because it exhibits a Lebesgue, not borel, measurable set @Crostul $\endgroup$ – jJjjJ Jan 7 '17 at 15:17

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