Tuesday, 18 December 2012

measure theory - Fast L1 Convergence implies almost uniform convergence



nN||fnf||1< implies fn converges almost uniformly to f, how to show this?




EDIT: Egorov's theorem is available. I have been able to show pointwise a.e. convergence using Chebyshev and Borel-Cantelli, I am having trouble trying to pass to almost uniform convergence using the absolute summability condition...


Answer



Put gn:=|fnf|, and fix δ>0. We have nNgnL1< so we can find a strictly increasing sequence Nk of integers such that nNkgn1δ4k. Put Ak:={xX:supnNkgn(x)>21k}. Then AknNk{xX:gn(x)2k} so
2kμ(Ak)nNk2kμ{xX:gn(x)2k}nNkgn1δ4k,


so μ(Ak)δ2k. Put A:=k1Ak. Then μ(A)k1μ(Ak)δk12k=δ, and if xA we have for all k: supnNkgn(x)21k so supnNksupxAgn(x)21k. It proves that gn0 uniformly on Ac, since for a fixed ε>0, we take k such that 21k, so for nNk we have supxAgn(x)ε.


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