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Let $f$ be defined on a measurable set $E \subset \mathbb R^n$. How to show that if $\{a<f<+\infty\}$ and $\{f=-\infty\}$ are measurable for every finite $a$, then $f$ is measurable? I think I need to separate the set $\{a<f<+\infty\}$, then do some set operations. Any hint?


EDIT: This question is part of a corollary.enter image description here

The definition of measurable function is that $f$ is called a Lebesgue measurable function on $E$, or simply a measurable function, if for every finite $a$, the set $$\{\mathbf{x}\in E: f(\mathbf{x})>a \}$$ is a measurable subset of $\mathbb R^n$.

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  • $\begingroup$ Isn't it $\{f=+\infty \}$ measurable instead of $\{f=-\infty \}$ ? Moreover, what is your definition of measurable function ? $\{-\infty \leq f <a\}$ measurable for all $a$ ? $\endgroup$ – Surb Jul 14 '18 at 15:53
  • $\begingroup$ @Surb I have posted a corollary and a definition of measurable function. $\endgroup$ – user398843 Jul 14 '18 at 16:23
  • $\begingroup$ @Yanko Does it really duplicate the other post? $\endgroup$ – user398843 Jul 14 '18 at 16:51
  • $\begingroup$ Oh I see, the main difference is the infinity part. I'll see if I can remove that close vote. $\endgroup$ – Yanko Jul 14 '18 at 17:55
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It is enough to show that the set $\{x\in E: \ f = +\infty\}$ is measurable, since then for each finite $a$ we get that the set $$ \{x \in E: \ f(x) > a \} = \{x\in E: \ f = +\infty\} \cup \{x \in E: \ +\infty>f(x) > a \} $$ is a union of two measurable sets, hence measurable itself.

To see that $\{x\in E: \ f = +\infty\}$ is measurable, observe that in view of the measurability of $\{x\in E: \ f = -\infty\}$ its complement is also measurable, hence $$ \tag{1} \{x\in E: \ f = -\infty\}^c = \{x\in E: \ f > - \infty\} = \\\{x\in E: \ f = +\infty\} \cup \left( \bigcup\limits_{n=-\infty}^{+\infty} \{x \in E: \ n<f(x) < +\infty \} \right). $$ Now the union in big brackets is measurable as a countable union of measurable sets. Since the left-hand side of $(1)$ is also measurable, the claim follows.

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