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Let $\{f_n\}_{n=1}^{\infty}$ be a sequence of real valued measurable functions on $[0,1]$. Show that there is a sequence of positive real numbers $\{a_n\}_{n=1}^{\infty}$ such that $a_nf_n \rightarrow 0$ a.e. on $[0,1]$.

Here is my idea. Let $E=[0,1]$. Let $F_n \subseteq [0,1]$ and closed such that $m(E\setminus F_n)<\epsilon$. Also, let $g_n$ be a continuous function on $F_n$, and thus bounded since $F$ is closed. Each $g_n$ is bounded by a corresponding $M_n$.

My goal was to use Lusin's theorem, but I'm starting to think that my idea is a dead end.

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    Maybe it will help to note that for each measurable $f$ as in the problem statement and each $\epsilon > 0$ there's some $N$ such that the set $\{x : |f(x)| > N\}$ has measure less than $\epsilon$.2011-09-29
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    Sorry, I'm not sure if I follow.2011-09-29
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    Look at $N^{-1}f$: $\vert N^{-1}f(x)\vert\le 1$ except on a set of measure less than $\epsilon$. $N^{-2}f$ does even better: $\vert N^{-2}f(x)\vert\le 1/N$ except on a set of measure less than $\epsilon$.2011-09-29

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