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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 $f_n(x)$?

Thanks!

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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. –  беркай 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! –  беркай Dec 18 '12 at 13:41
    
This hypothesis is irrelevant--but no big deal. –  Did Dec 18 '12 at 14:56
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1 Answer

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).

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