Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $\bigcap\limits_{n=1}^\infty\bigcup\limits_{k=n}^\infty A_k$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.