# Sequence of measurable functions such that $a_nf_n \rightarrow 0$

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.

• 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 $\epsilon$. – user83827 Sep 29 '11 at 3:11
• Sorry, I'm not sure if I follow. – wrldt Sep 29 '11 at 3:16
• 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 $\epsilon$. – Brian M. Scott Sep 29 '11 at 3:25

You could use Lusin's theorem to show that a real valued measurable function $f$ is bounded off a set of arbitrarily small measure, but you do not need to do so. Another way is to note that $[0,1]=\bigcup\limits_{n=1}^\infty f^{-1}(-n,n)$ implies $\lim\limits_{n\to\infty}m(f^{-1}(-n,n))=1$.
This implies that for each $n$ and each $\delta_n>0$, there exists $M_n>0$ such that $A_n=\{x:|f_n(x)|>M_n\}$ has measure less than $\delta_n$. You can choose the sequence $(\delta_n)$ to ensure that $m(A_n\cup A_{n=1}\cup A_{n+2}\cup A_{n+3}\cup\cdots)$ goes to zero. You can then choose a sequence $(a_n)$ such that $(a_n f_n)$ converges to zero pointwise off of the set $\bigcap\limits_{n=1}^\infty\bigcup\limits_{k=n}^\infty A_k$.
• would you please elaborate the part where you are saying "You can choose a sequence $(a_n)$" such that $(a_n f_n)$ converges to zeero pointwise" – Deepak Jan 2 '13 at 21:33
• @Deepak, If $x$ is not in $\bigcap\limits_{n=1}^\infty\bigcup\limits_{k=n}^\infty A_k$, then for sufficiently large $n$, $|f_n(x)|\leq M_n$. Thus, for example, $a_n=\dfrac{1}{nM_n}$ suffices. – Jonas Meyer Jan 2 '13 at 22:13