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Let $f:\mathbb R\to\mathbb R$ a measurable function. prove or disprove that $E=\{(x,\alpha)\mid 0\leq \alpha< |f(x)|\}$ is measurable.

I know it's measurable, but I really have no idea how to prove it.

P.S: Does $E=\{(x,\alpha)\mid 0\leq \alpha\leq |f(x)|\}$ still measurable or not ?

Rick
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    I don't get the definition of $E$. What do you mean by $(x,\alpha)$, is that a ordered pair or is that an open interval, or something else? – user160738 Nov 09 '15 at 20:12
  • This is all the problem of the notation $(a,b)$ for ${x\mid a<x<b}$, we can have confusion between ordered pair or open interval. French notation $]a,b[$ don't give this confusion. But here it's an ordered pair (I don't know what we could do with open interval here) – Surb Nov 09 '15 at 20:23
  • Related: http://math.stackexchange.com/questions/1496524 – Watson Jun 21 '16 at 15:46

1 Answers1

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1) Let $Q$ measurable. If $f=\boldsymbol 1_Q$, then $E=(Q^c\times \{0\})\cup(Q\times [0,1])$ which is measurable.

2) If $f$ measurable, there is a sequence $\{\varphi_n\}$ s.t. $\varphi_n\nearrow |f|$. Let $$E_n=\{(x,\alpha)\mid 0\leq \alpha < \varphi_n\}.$$ We have that $E_{n}\subset E_{n+1}$ and that $$E=\bigcup_{i=1}^\infty E_n.$$ By 1) all the $E_n$ are measurable, and thus $E$ is measurable.

For your PS, the answer is yes.

Surb
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