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 Subject: modes of convergence Category: Science > Math Asked by: madukar-ga List Price: \$5.00 Posted: 27 Feb 2003 12:20 PST Expires: 29 Mar 2003 12:20 PST Question ID: 167947
 ```hi dannidin-ga could you please help me in this problem: give an example showing that "convergence in measure' of a sequence {fn}running from n=1 to infinity subset of M does not imply convergence of {fn(X)} running from n=1 to infinity at any point x belongs to X. So, in particular convergence in measure does not imply convergence almost everywhere. also give an example of a sequence {fn} running from n=1 to infinity subset of L' which converges in measure to a function f belongs to L' but such that not even a subsequence of {fn}running from n=1 to infinity is convergent in L'.```
 ```Hi Madukar-ga, For your first question, believe it or not but I already answered it! Check out https://answers.google.com/answers/main?cmd=threadview&id=120524 The sequence of functions that I construct there (in part b.) converges in measure to 0, since the set of points where the function fn is not zero has measure 1/2^k, where k is approximately log(n) (to base 2). However, for EVERY point x in [0,1], fn(x) is not a convergent sequence of real numbers. Now for the second question: Define the sequence of functions gn:[0,1]->R by gn(x) = 0 if x > 1/2^n = 2^n if x < or = 1/2^n Then gn converges in measure to 0 (for the same reason as stated above for fn). But gn cannot converge in L' to 0, and not even any subsequence can converge to 0 in L', because for all n we have !!gn!!1 = !!gn - 0!!1 = 1 (!!gn - 0!! measures the distance in L' between gn and 0. If this does not converge to zero then gn does not converge in L' to 0) Hope this is clear, if not please ask and I'll clarify. Regards, dannidin```