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Math Help - Lebesgue Integration

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    Smile Lebesgue Integration

    Let (S,\Sigma,\mu) be a measure space. Prove that if f is a Lebesgue integrable function, then for each \varepsilon > 0 there exists a \delta > 0 such that \int_A \mid f \mid \mathrm{d} \mu for each A \in \Sigma with \mu(A) < \delta.
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    Quote Originally Posted by syme.gabriel View Post
    Let (S,\Sigma,\mu) be a measure space. Prove that if f is a Lebesgue integrable function, then for each \varepsilon > 0 there exists a \delta > 0 such that \int_A \mid f \mid \mathrm{d} \mu \color{red}<\varepsilon for each A \in \Sigma with \mu(A) < \delta.
    Suppose on the contrary that for some \varepsilon>0 there exists, for each n=1,2,3,..., E_n\in\Sigma such that \mu(E_n)<2^{-n} but \int_{E_n}|f|\,d\mu\geqslant\varepsilon. Let E = \limsup_nE_n = \bigcap_nF_n, where F_n = \bigcup_{k=n}^\infty E_k. Then \mu(E)\leqslant\sum_{i=n}^\infty\mu(E_i)<2^{-n+1} for all n, and hence \mu(E)=0.

    On the other hand it follows from Dominated Convergence or some such theorem that \int_E|f|\,d\mu = \lim_n\int_{F_n}|f|\,d\mu \geqslant\limsup_n\int_{E_n}|f|\,d\mu\geqslant\var  epsilon. That is a contradiction since clearly |f| cannot have a nonzero integral over a null set.

    The idea of this proof comes from the fact that if we define a new measure \nu on \Sigma by \nu(A) = \int_A|f|\,d\mu\ (A\in \Sigma), then \nu\ll\mu. There is then a standard theorem (Theorem 30B in Halmos's Measure Theory) saying that absolutely continuous measures have this "continuity" property. I just took Halmos's proof and re-wrote it for this situation.
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