Its clear that convergence in $L^2$ does not necessarily imply pointwise convergence a.e but does this work the other way too? I considered the characteristic function $\chi_{[n, n+1]}$ which I guess pointwise converges to 0 in $L^2$ but not for $x \in (0,1)$ I think but I want the example for the other way around for something that converges pointwise but not in $L^2$ if the function in $L^2$ is bounded?
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1You may want to look at the dominated convergence theorem and Riesz-Scheffé lemma. – nejimban Dec 05 '22 at 04:06
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It is unclear to me what you are asking. First of all are you interested in $L^2(0,1)$ for all of your questions? Secondly what do you mean by $\lim_{n\to \infty}|f_n|2=|f_n|_2<\infty$? Are you actually asking whether, if $f_n\to f$ a.e and $\lim{n\to \infty}|f_n|_2=|f|_2<\infty$ then $f_n\to f$ in $L^2$? – K.Power Dec 05 '22 at 04:06
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sorry for not being clear I am asking if fn -> f a.e. in the measure space (0,1) and if ||fn|| in L2 is bounded by the sup then does it also converge? im wondering about two cases 1. if it converges if it is bounded by some constant C and 2. if it converges in general if the ||fn|| < infty sorry and thanks so much – Dec 05 '22 at 04:13
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If I am understanding your question properly, then the answer is no. Consider the sequence of indicator functions $f_n:=\sqrt{n}\mathbb 1_{(0,1/n)}$. This sequence converges $a.e$ to the $0$ function on $(0,1)$, but $$\|f_n\|_2^2=n\int_{(0,1)}\mathbb 1_{(0,1/n)}(x)dx=1.$$ This is a counterexample to both of your questions as elucidated in your comment on your question. Note that this does not contradict the dominated convergence theorem, because $(f_n)$ is not bounded above a.e. (I assumed this is what you meant by the uniform boundedness theorem). As pointed out in the comments by @nejimban, if $f_n\to f$ a.e., and $\|f_n\|_2\to \|f\|_2$, then $\|f_n-f\|_2\to 0$. See here for a proof.

K.Power
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[+1] Also note that $|f_n|_2\not\to|f|_2$ in this example, so it does not contradict Riesz-Scheffé lemma either. – nejimban Dec 05 '22 at 05:34