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 Subject: measure theory Category: Science > Math Asked by: madukar-ga List Price: \$10.00 Posted: 06 Dec 2002 14:11 PST Expires: 05 Jan 2003 14:11 PST Question ID: 120521
 ```let (X,M,mu)be a measure space with mu(X) < infinity, (fn) n belongs to N a sequence of bounded measurable functions fn : X to R, f :X to R with fn converges to f uniformly on X. (a) show that integral fn converges to integral f. (b) construct a counterexample showing that the assumption mu (X) < infinity cannot be dropped. (c) construct a counterexample showing that the assumption of uniform convergence cannot be relaxed to pointwise convergence. hint: verify that the function fn are unifromly bounded,then apply the dominated onvergence theorm.```
 ```Hi again, I did not understand your hint. It sounds like it applies to (a), but I find it much easier to prove (a) without using a "heavy" theorem like the dominated convergence theorem: (a) I use the triangle inequality for integrals, which says that |integral(f)| < integral(|f|) (here and below, for convenience I use "<" to mean "< or =") (If you don't know the proof of this, tell me and I'll write it for you.) Anyway, we need to show that integral(fn)-->integral(f), or |integral(fn)-integral(f)|-->0. But |integral(fn)-integral(f)| = |integral(fn-f)| < integral(|fn-f|) < < mu(X)*sup(|fn(x)-f(x)|) where the supremum is over all x in X. |fn-f| converges uniformly, so for any e > 0 (e=epsilon) there is an N such that for all n>N, we have |fn(x)-f(x)|N |integral(fn)-integral(f)| < mu(X)*e Since e>0 is arbitrary this proves that integral(fn)--->integral(f). (here I am using the fact that mu(X)0 uniformly, but integral(fn) = infinity --/-> integral(0)=0. (c) Take (X,M,mu)=([0,1],B=borel sets,mu=Lebesgue measure on [0,1]). Take fn = n*indicator function of (0,1/n). Then fn-->0 pointwise for any x (for x=0, fn(x)=0 for all n; for x>0 fn(x)=0 for any n>1/x), but integral(fn) = 1 (for all n) --/-> integral(0)=0 Regards, dannidin``` Request for Answer Clarification by madukar-ga on 08 Dec 2002 14:00 PST ```hi, could you please prove the theorm by verify that the given function fn is uniformly bounded and by dominated convergence theorm. waiting for your reply.``` Clarification of Answer by dannidin-ga on 09 Dec 2002 00:23 PST ```Sure I can: f is a bounded function, let's say |f|N we have |fn-f| < 1 for all x in X. Therefore, for all n>N we have |fn|< M + 1 by the triangle inequality. So the sequence fn is uniformly bounded (you can have it uniformly bounded even for all values of n by replacing M+1 by a bigger constant that also bounds f1, f2, ..., fN). Now we can apply the dominated/bounded convergence theorem to say that integral(fn)--->integral(lim(fn))=integral(f) as n->infinity. Note that we can only use the dominated convergence theorem because mu(X)
 madukar-ga rated this answer: `outstanding answer.`