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Let $f_n$ be a sequence of measurable functions on $\mathbb R$.

Show that the set $A:= \{x \in \mathbb R \mid f_n > 0\text{ for infinitely many } n \}$ is measurable.

If I write the set $A$ like intersection of the union of a measurable set, then I am done.

But I can not. Please help me.

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  • $\begingroup$ There exists an infinity of n translates by an intersection on k, of an union on $n\geq k$. That gives $\bigcap_{k=1}^{\infty} \bigcup_{k\geq n} ...$ $\endgroup$ – Jean-Claude Colette Jan 19 at 18:50
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Let $$E_n:= \{f_n > 0\}=\{x\in\mathbb R\mid f_n(x)>0\}$$

$$A = \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty E_n$$

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For any finite $A\subset \mathbb{N}$ let $P_A = \{x\in\mathbb{R}|f_n(x) > 0\iff n\in A\}$. This set is measurable as it is a finite intersection of measurable sets of the form $f_n^{-1}((-\infty,0])$ for $n\notin A$ etc. Now take the union of all $P_A$. Since $A$ are finite there are only countably many of the so that the union is also measurable. However, this is exactly the set of all points which are positive on just finitely many of the $f_n$ so it's complement, which is also measurable, is the set you were looking for.

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