# Show different limits under different mode of convergence equals almost everywhere

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 $m(|f_n-f_2|\gt\epsilon)\leqslant\epsilon^{-2}\|f_n-f\|_2^2$. – Did Aug 25 '12 at 15:17
@Norbert Asked 5 hours ago is a bit soon for a question to be declared unanswered, don't you think? – Did Aug 25 '12 at 20:10
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 comments – Norbert Aug 25 '12 at 22:24
@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. – Did Aug 25 '12 at 22:48
@did So what you don't like? – Norbert Aug 25 '12 at 23:49

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.