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Let $(X,\mathcal{M},\mu)$ be a measure space. Show that if $f\in L^+$ (measurable and non-negative functions) and $\int fd\mu<\infty$ then, $\forall \varepsilon >0$ $\exists E\in\mathcal{M}$ s.t. $\mu(E)<\infty$ and $\int_E fd\mu>\int f d\mu-\varepsilon$

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Let $E_k = \{x \in X : f(x) > k^{-1}\}$. Then $\mu(E_k) < \infty$ by Chebyshev's inequality, and $$\int f \, d\mu = \int_{\cup E_k} f \, d\mu\ = \lim_{k \to \infty} \int_{E_k} f \, d\mu$$ by e.g. the dominated convergence theorem.

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  • $\begingroup$ I tried to write it down but i am not sure if the conclusion is right ($\chi$ is the indicator function): Define the sequence $$ f_k:=f\chi_{E_k} $$ where $f_k\leq f_{k+1}$ and given that $0\leq f_k\leq f\in L^+$ apply MCT and $$ \lim_{k\rightarrow \infty}\int_{E_k} f=\int f. $$ Given that $E_k\subset E_{k+1}$ $\forall \varepsilon>0 \exists E_\varepsilon\in\{E_k\}$ s.t. $\mu(E_\varepsilon)<\infty$ and $$ \int_{E_\varepsilon} f\geq\int f-\varepsilon. $$ $\endgroup$ – Rgkpdx Dec 11 '13 at 16:22

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