let be a Lebesgue integrable function on [0,1], and let be a sequence of bounded measurable functions on [0,1] that converges uniformly to a function . Prove that

.

help will be appreciated so much

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- December 21st 2009, 07:53 PMGTOLebesgue integral
let be a Lebesgue integrable function on [0,1], and let be a sequence of bounded measurable functions on [0,1] that converges uniformly to a function . Prove that

.

help will be appreciated so much - December 22nd 2009, 12:04 AMGTOdominated convergence theorem
i dont see why has to converge uniformly to . to me if , then (i am not sure how to show). and since there exists such that and since is integrable, so is .

by dominated convergence theorem, i think we get what we want.

can someone help me with this one? - December 22nd 2009, 02:17 AMSwlabr
Firstly, you will want to use the monotone convergence theorem. This gives us equality (your working with the Lebesgue integral and your lecturer is trying to show you how inadequate the Riemann integral is; the monotone convergence theorem doesn't work for Riemann, but it does for Lebesgue so they will tend to use it lots). You can use this here as you are in a compact interval and so a uniformly convergent sequence is non-decreasing.

So, can you prove that your sequence uniformly converges to if the sequence uniformly converges to ? - December 22nd 2009, 02:37 PMGTOmonotone convergence thm
doesnt a sequence has to be nonnegative to use monotone convergence thm?

- December 22nd 2009, 05:49 PMputnam120
You can also use the dominated convergence theorem once you show that is bounded (which it is). I assume that when you say is integrable you mean that .

Also at Swlabr how do you know that the sequence non-decreasing? - December 23rd 2009, 02:31 AMSwlabr
I'm now a tad confused. I was sure I found a result which said that any uniformly convergent sequence in a compact space is non-decreasing. However, I cannot seem to find this now and in retrospect it doesn't seem to make that much sense - something can converge uniformly from above...

- January 9th 2010, 12:04 AMGTO
- January 9th 2010, 06:01 PMputnam120
Let , since uniformly there exist N such that for . Thus . Now choose .

Now , so you can use DCT.