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Suppose that a sequence of bounded and continuous functions $f_n$ converges uniformly to $f_1$ and $f_n$ converges to $f_2$ in $L^2$ sense, then how to show $f_1= f_2$ a.e.?

I tried the following: let $A_\epsilon = \{x:|f_1(x)-f_2(x)|>\epsilon\}$, then $m(A_\epsilon) < m(|f_n - f_1|>\epsilon) + m(|f_n - f_2|>\epsilon)$. Let $n$ go to infinity, then the first part of RHS goes to zero by uniform convergence, but I cannot do anything to $L^2$-convergence.

Can anyone show me how to solve this question? Thanks in advance .

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    You are on the right track: use $m(|f_n-f_2|\gt\epsilon)\leqslant\epsilon^{-2}\|f_n-f\|_2^2$.2012-08-25
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    @Norbert *Asked 5 hours ago* is a bit soon for a question to be declared *unanswered*, don't you think?2012-08-25
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    I think people shy to post answer that you have already gave in first comment. I really don't like common practice of posting answers as comments2012-08-25
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    @Norbert What you say does not correspond to my experience, as I have seen countless examples of the opposite happening on this site. Anyway, your second comment forces me to interpret your first one quite differently than I first did, and in a way which *I really don't like*.2012-08-25
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    @did So what you don't like?2012-08-25
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    @Norbert Say, would you be trying to troll me?2012-08-25
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    No of course I don't. I really don't understand why you often post answers as comments.Well nevermind, I see that this talk is wasting of reverences.2012-08-26
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    Cathais09 Did you manage to transform the hint in my first comment into a full proof or do you wish for some more detailed indications?2012-08-27

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