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 Subject: integrationof complex functions Category: Science > Math Asked by: madukar-ga List Price: \$10.00 Posted: 06 Dec 2002 14:21 PST Expires: 05 Jan 2003 14:21 PST Question ID: 120524
 ```let (fn)where n belongs to N be a sequence of Lebesgue measurable functions fn : [0,1]to [0,1]. (a) prove that if fn tends to 0 a.e, then integral fn over [0,1] tends to 0. (b) show that the reverse implicatoin is false by constructing a sequence of Lebesgue meaurable functions fn :[0,1] to [0,1] with integral fn over [0,1] such that (fn)where n belongs to N does not converge at any point x belongs to [0,1].```
 ```Hi Madukar-ga, (a) fn is a sequence of bounded measurable functions that converges to 0 a.e., therefore by the bounded convergence theorem lim (integral fn) = integral(lim fn) = integral(0) = 0 (b) Define a sequence of indicator functions of intervals as follows: I[a,b] denotes the indicator/characteristic function of the interval [a,b], i.e. the function that is equal to 1 on [a,b] and 0 elsewhere. f0 = I[0,1] f1 = I[0,1/2] f2 = I[1/2,1] f3 = I[0,1/3] f4 = I[1/3,2/3] f5 = I[2/3,1] f6 = I[0,1/4] f7 = I[1/4,1/2] f8 = I[1/2,3/4] f9 = I[3/4,1] f10 = I[0,1/5] f11 = I[1/5,2/5] ... fn = I[k/m,(k+1)/m] if n = m(m-1)/2 + k Then integral(fn) = 1/m (for n=m(m-1)/2+k) ---> 0 as n->infinity. However, clearly for any fixed x in [0,1], the sequence of number fn(x) cannot converge, because there are infinitely many values of n for which fn(x) is zero (any n such that x is not contained in the interval [k/m,(k+1)/m]) and infinitely many values of n for which fn(x) is 1 (any n such that x IS in the abovementioned interval). (A professor of mine used to call this example "the Goodyear blimp", because the graphs of the function sequence fn look like a blimp that passes back and forth in the sky...) Hope this helps, thanks for the question. I'm here if you need any clarifications. Cheers, dannidin```
 madukar-ga rated this answer: `good it really helped me a lot`