# Real sequences and convergence almost everywhere.

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.

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 $$a_{n_j}=\cdots=a_{n_{j+1}-1}=j$$.
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 $$p=a$$).