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So i was looking at a question on stack overflow about almost sure convergence and l2 convergence. Im not sure if i understand it exactly. The question is this. Why does $L^2$ convergence not imply almost sure convergence

Op asked if l2 convergence of a sequence of functions gurantees almost sure convergence and the answer was no because you could have l2 convergence and not pointwise convergence.

The example given was the type writer sequence. It converges in l2 to 0 but does not converge to 0 almost surely. This is because it doesn't agree with 0 on the entire interval of [0,1], every value will take on both 0 and 1 values infinitely often.

The reason the box is moving is because if it was fixed at one edge like the retracting rectangle f_n=1 if x is in [0,1/n] and 0 elsewhere converges in l2 to 0 and is also almost surely converging to 0 because lim[0,1/n] has measure 0.

Am i understanding this right?

Qwertford
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    If you take all intervals of the type $[0,a_n]$ then $L^{2}$ convergence implies $a_n \to 0$ which implies that $f_n=I_{[0,a_n]} \to 0$ almost everywhere. If this is what you are thinking then you are right. – Kavi Rama Murthy Aug 07 '19 at 23:19
  • Yeah that was. And thats the reason the counter example to l2 convergence implies almost sure convergence was that fn and 0 disagree on the whole of [0,1] right? – Qwertford Aug 07 '19 at 23:20
  • You are on the right track. – Kavi Rama Murthy Aug 07 '19 at 23:22

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