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Math Help - Need to prove |f(x)|<= 1 almost every where.

  1. #1
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    Need to prove |f(x)|<= 1 almost every where.

    Let f :X \rightarrow R be a measurable function in X with m(X)< \infty.
    Show that if f^n is Lebesgue integrable for every n and that \lim_{n\to\infty}\int f^n\,dm exists in R,then -1\le f(x)\le 1 almost every where.

    Can anyone help me with this problem??
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  2. #2
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    Here is just a general idea. Assume that |f|>1 on some measurable set with m(U)>0. Then f^n>M for any M\in\mathbb{R} once n is large enough. So what will this do to \int f^ndm?

    Hint: \int f^ndm=\int_U f^ndm+\int_{U^c}f^ndm.
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  3. #3
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    Just to elaborate a bit on putnam120's suggestion, for k=1,2,3,..., let S_k = \{x\in X: |f(x)|\geqslant 1+1/k\}. Then \int_X\!|f^n|\,dm\geqslant \int_{S_k}\!\!|f^n|\,dm\geqslant (1+1/k)^nm(S_k). If m(S_k)\ne0 then this goes to infinity as n\to\infty. Also, if n is even then |f^n| = f^n, so \lim_{n\to\infty}\int_X\!f^n\,dm will not exist in that case.

    Therefore m(S_k) = 0 for all k and hence \textstyle m\left(\bigcup_{k}S_k\right) = 0. But \textstyle \bigcup_{k}S_k = \{x\in X:|f(x)>1\}. Therefore -1\leqslant f(x)\leqslant 1 almost everywhere.
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