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:sup. Then A_k\subset\bigcup_{n\geq N_k}\left\{x\in X: g_n(x)\geq 2^{-k}\right\} so
2^{-k}\mu(A_k)\leq \sum_{n\geq N_k}2^{-k}\mu\left\{x\in X: g_n(x)\geq 2^{-k}\right\}\leq \sum_{n\geq N_k}\lVert g_n\rVert_1\leq \delta 4^{-k},
so \mu(A_k)\leq \delta 2^{-k}. Put A:=\bigcup_{k\geq 1}A_k. Then \mu(A)\leq \sum_{k\geq 1}\mu(A_k)\leq \delta\sum_{k\geq 1}2^{-k}=\delta, and if x\notin A we have for all k: \sup_{n\geq N_k}g_n(x)\leq 2^{1-k} so \sup_{n\geq N_k}\sup_{x\notin A}g_n(x)\leq 2^{1-k}. It proves that g_n\to 0 uniformly on A^c, since for a fixed \varepsilon>0, we take k such that 2^{1-k}, so for n\geq N_k we have \sup_{x\notin A}g_n(x)\leq\varepsilon.


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