Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Assume that we are given a sequence of continuous functions $f_n(x)$ on $[0,1]$.

How to show the existence of a sequence $a_n$ and a set $A$ with $\mu(A^c)=0$ so that

$$ \lim_{ n\to \infty} \frac{f_n(x)}{a_n}=0, ~~ \forall x\in A. $$

I choose a sequence $a_n$ such that $ \mu (\phi_n) \leq 1/2^n$ where $$ \phi_n := \left\{ x: \frac{|f_n(x)|}{a_n} \geq \frac{1}{n} \right\}. $$

Since $\sum_n \mu (\phi_n) < \infty$, using Borel-Cantelli Lemma we have $\mu(\limsup_n \phi_n)=0$.

It seems okay if we say that $\limsup_n \phi_n = A^c$. How can we write it clearly in full details? Also, how can we assure the existence of $a_n$, how to construct such a sequence by means of $f_n(x)$?


share|cite|improve this question
This assumes that each function $f_n$ is finite almost everywhere. – Did Dec 18 '12 at 12:30
Yes, that's correct due to Extreme Value Theorem. – Ragnar Dec 18 '12 at 12:54
?? EVT has nothing to do here, none of these functions is assumed to be continuous. – Did Dec 18 '12 at 13:18
oh, i forgot to say that $f_n(x)$ are continuous, thank you! – Ragnar Dec 18 '12 at 13:41
This hypothesis is irrelevant--but no big deal. – Did Dec 18 '12 at 14:56

We have for almost every $x\in A$ that there exists $N=N(x)$ such that for $n\geqslant N$, $\frac{|f_n(x)|}{a_n}\leqslant \frac 1n$ (hence $\frac{|f_n(x)|}{a_n}\to 0$).

The number $a_n$ exists because $\lim_{R\to \infty}\lambda\{|f_n|\gt R\}=0$ (we only need $f_n$ to be measurable).

share|cite|improve this answer

I'm a little confused by this question: if the functions $f_n$ are continuous, then we don't need any measure theory here at all. Simply set $a_n = n \sup_{x \in [0,1]} |f(x)|$ (where $a_n$ is finite because a continuous function on a compact set is bounded). Then $f_n(x)/a_n \to 0$ for every $x$, not just almost every $x$. (In fact, $f_n/a_n \to 0$ uniformly.)

share|cite|improve this answer

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.