Lebesgue measurable function

Asked Jun 16 '15 at 14:46

Active Jun 16 '15 at 16:57

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I'm reading an article and have to use the following statement.

If $g$ is a real valued function on $\mathbb{R}^{2}$ such that $g_{x}$ is Lebesgue measurable for all $x \in E$ and $g^{y}$ is continuos for all $y \in \mathbb{R}$ then $g$ is Lebesgue measurable.

Is it true? Does anyone know why? I tried to justify and could not.

edited Jun 16 '15 at 15:19

zoli

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asked Jun 16 '15 at 14:46

Craneviter

An idea for a place to start: given an open set $U$ in $\mathbb{R}$, notice that $\bigcup_{q \in \mathbb{Q}} g_q^{-1}(U)$ is measurable, because it is a countable union of measurable sets. The same is true for the other section (continuity implies measurability). Can you somehow relate $g^{-1}(U)$ back to one of these sets? – Ian Jun 16 '15 at 15:33
Could you link the article? Could you explain the notation ($g_x$? $E$?)? – masterxilo Jun 16 '15 at 16:03
1

@Craneviter Please look here: http://math.stackexchange.com/questions/647235/counterexample-to-measurable-in-each-variable-separately-implies-measurable – Jun 16 '15 at 16:44
You can also have a look at this answer in a related Math.SE question. It gives a reference for a proof of your question (Lemma 4.51 of Infinite Dimensional Analysis (3rd Ed.), Aliprantis and Border). – Clement C. Jun 16 '15 at 19:14

Lebesgue measurable function

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