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Math Help - Lebesgue integral

  1. #1
    GTO
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    Lebesgue integral

    let f be a Lebesgue integrable function on [0,1], and let \{g_n\} be a sequence of bounded measurable functions on [0,1] that converges uniformly to a function g. Prove that
    \int_{[0,1]}g_nfdx \rightarrow \int_{[0,1]}gfdx.

    help will be appreciated so much
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  2. #2
    GTO
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    dominated convergence theorem

    i dont see why g_n has to converge uniformly to g. to me if g_n \rightarrow g, then g_nf \rightarrow gf(i am not sure how to show). and |g_nf| \leq Mf since there exists M such that |g_n| \leq M and since f is integrable, so is Mf.
    by dominated convergence theorem, i think we get what we want.

    can someone help me with this one?
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by GTO View Post
    let f be a Lebesgue integrable function on [0,1], and let \{g_n\} be a sequence of bounded measurable functions on [0,1] that converges uniformly to a function g. Prove that
    \int_{[0,1]}g_nfdx \rightarrow \int_{[0,1]}gfdx.

    help will be appreciated so much
    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 g_nf uniformly converges to gf if the sequence g_n uniformly converges to g?
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  4. #4
    GTO
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    monotone convergence thm

    doesnt a sequence has to be nonnegative to use monotone convergence thm?
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  5. #5
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    You can also use the dominated convergence theorem once you show that g is bounded (which it is). I assume that when you say f is integrable you mean that \int_0^1|f|d\mu<\infty.

    Also at Swlabr how do you know that the sequence non-decreasing?
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by putnam120 View Post
    You can also use the dominated convergence theorem once you show that g is bounded (which it is). I assume that when you say f is integrable you mean that \int_0^1|f|d\mu<\infty.

    Also at Swlabr how do you know that the sequence non-decreasing?
    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...
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  7. #7
    GTO
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    Quote Originally Posted by putnam120 View Post
    You can also use the dominated convergence theorem once you show that g is bounded (which it is). I assume that when you say f is integrable you mean that \int_0^1|f|d\mu<\infty.

    Also at Swlabr how do you know that the sequence non-decreasing?
    since g_n \rightarrow g uniformly, for \epsilon >0, there is N such that for all n \geq N and for all x \in [0,1], we have |g-g_n|< \epsilon.
    So |g|<|g_n|+\epsilon \leq M_n + \epsilon for all n \geq N. so g is bounded. but how can i used the dominated convergence theorem?
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  8. #8
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    Let \epsilon>0, since g_n\to g uniformly there exist N such that |g_n-g_N|<\epsilon for n>N. Thus |g_n|\le\epsilon+|g_N|\le\epsilon+M_N. Now choose M=\max(M_1,\dots,M_{N-1},M_N+\epsilon).

    Now |g_nf|\le M|f|, so you can use DCT.
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