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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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    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

1 Answers 1

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Markov's inequality does the job: we get that for each $\varepsilon>0$, $\lambda\{x,|f_n(x)-f_2(x)|>\delta\}\leqslant \frac 1{\varepsilon^2}\lVert f_n-f_2\rVert_{L^2}^2,$ hence following the notations in the OP, we get $\lambda(A_{2\varepsilon}\cap [-N,N])\leqslant \lambda(\{x,|f_n(x)-f_1(x)|>\varepsilon\}\cap [-N,N])+\frac 1{\varepsilon^2}\lVert f_n-f_2\rVert_{L^2}^2,$ hence $\lambda(A_{2\varepsilon}\cap [-N,N])\leqslant 2N\cdot \left[\sup_{[-N,N]}|f_n-f_1|>\varepsilon\right]+\frac 1{\varepsilon^2}\lVert f_n-f_2\rVert_{L^2}^2,$ where $[P]$ is when when $P$ is satisfied and $0$ otherwise.

When $N$ and $\varepsilon$ are fixed, the RHS goes to $0$ as $n$ goes to infinity. Hence, $\lambda(A_{2\varepsilon}\cap [-N,N])=0$ for all $N$ and $\varepsilon$, giving the wanted conclusion.