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I'm aware of the definition of the measurable function. But I was wondering how to prove simple function is measurable? It would be better have some detailed proof.

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Consider a simple function $$ f = \sum_{i=1}^na_i1_{E_i}, $$ where $E_1,\ldots,E_n$ are disjoint measurable sets. To see that $f$ is measurable, it suffices to show that $$ f^{-1}(-\infty,a] $$ is measurable for every $a\in\mathbb{R}$. Given $a\in\mathbb{R}$, note that $$ f^{-1}(-\infty,a] = \bigcup\{E_i \mid i\in\{1,\ldots,n\}\ \text{such that}\ a_i\le a\} $$ is a union of measureable sets. Therefore $f$ is measurable.

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  • $\begingroup$ It could happen that $x \in D =\left(\bigcup_{i = 1}^{n} E_{i}\right)^{c}$. In this case we have that $f(x) = 0$. If $a \geq 0$, we would have $f^{-1} (- \infty, a] = \cup \{E_{i} \mid i \in \{1, \dots, n\} \text{ such that } a_{i} \leq a\} \cup D$. $\endgroup$
    – DIEGO R.
    Sep 29, 2021 at 21:20

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