Almost everywhere equal implies measurable

Asked Nov 08 '19 at 14:03

Active Nov 08 '19 at 14:03

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I am still new to Lebesgue measure theory. I would like to get a verification on my proof.

Statement: Assume a complete measure space. If $f$ is a measurable function and $f=g$ almost everywhere, then $g$ is measurable.

Proof: Let $c$ be a real number and $N$ the null set such that $f=g$ on $N^C$. Hence, $$g^{-1}(c,\infty)=[g^{-1}(c,\infty)\cap N]\cup[g^{-1}(c,\infty)\cap N^C].$$

The first RHS bracket is measurable since it is the subset of $N$ and the space is complete. The second is measurable because it is equal to $f^{-1}(c,\infty)\cap N^C$, where $f^{-1}(c,\infty)$ is measurable by assumption and $N^C$ is measurable because $N$ is. Q.E.D.

Are there missing details in the proof? Thanks a lot.

asked Nov 08 '19 at 14:03

10understanding

3

You are right. Here you are assuming all functions are real valued not extended real valued. – Sumanta Nov 08 '19 at 14:05
Ah you are right. What if the images of f and g are extended? Does the proof still follow? – 10understanding Nov 08 '19 at 14:06
Wait, maybe it still follows but I must change my notation to ${x \mid f(x)>c}$ instead. How? – 10understanding Nov 08 '19 at 14:08
1

You have to replace $(c,\infty)$ by $(c,\infty]$. – Sumanta Nov 08 '19 at 14:08
In general to state function measurability, I just need to consider {f>c} for a (not necessarily extended) real number c, correct? – 10understanding Nov 08 '19 at 14:10

Almost everywhere equal implies measurable

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