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Let $(X,\mu,\mathcal{A})$ be a finite measure space and $f_n$ measurable functions such that $f_n \to 0 $ almost everywhere.

Show that exists a sequence $a_n \to +\infty$ auch that $a_nf_n \to 0$ a.e.

I managed (by using the Borel-Cantelli lemma) to find a subsequence $a_{n_m}$ such that $a_{n_m}f_{n_m} \to 0 $ a.e using the convergence in measure(since we have convergence a.e),but i could not solve it.

Can someone give me a hint?

I do not seek a full solution.

Thank in advance.

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Assuming that $\mu(X)<\infty$, $f_n\to 0$ a.e. iff for every $\epsilon>0$, $\mu(\sup_{k\ge n}|f_k|>\epsilon)\to 0$ as $n\to\infty$. In this case $\{a_n\}$ can be constructed as follows. Pick the sequence $n_j$ s.t. $\mu(\sup_{k\ge n_j}|f_k|>\epsilon/j)\le j^{-1}$ and for each $j\ge 1$, set $a_{n_j}=\cdots=a_{n_{j+1}-1}=j$.

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  • $\begingroup$ I edited...we are in a finite measure space $\endgroup$ – Marios Gretsas Jul 28 at 19:23
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This is a reply to the original version of the question, where we were given an arbitrary measure space. The example below shows it's not true in general; it appears to be true for a finite measure space, one conjectures that $\sigma$-finite is enough.

Let $X$ be the set of all real sequences $a=(a_n)$ such that $a_n\to\infty$. Say every subset of $X$ is measurable, and let $\mu$ be counting measure, so convergence almost everywhere is the same as convergence at every point.

For each $a=(a_n)\in X$ let $(f_n(a))$ be a sequence such that $f_n(a)\to0$ and $a_nf_n(a)\to\infty$. So $f_n\to0$ pointwise, but for every $a=(a_n)$ with $a_n\to\infty$ there exists $p\in X$ such that $a_nf_n(p)\to\infty$ (namely $p=a$).

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  • $\begingroup$ You are right..i edited..we are in a finite measure space $\endgroup$ – Marios Gretsas Jul 28 at 19:23

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