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Let $\Omega\subseteq\mathbb{R}^N$ be an open set and $f:\Omega\to[0,+\infty[$ a measurable function, bounded over each compact $K\subset\Omega$. If there is a $C>0$ such that $$\int_{K}f\operatorname{dm}\leq C$$ and a sequence $\{ K_j\}_{j\in\mathbb{N}}$ of compact sets such that $K_j\subset\operatorname{int}(K_{j+1})$ and $\bigcup_{j\in\mathbb{N}}K_j=\Omega$, how to prove that $$\lim_{j\to\infty}\int_{K_j}f\operatorname{dm} =\sup\left\{\int_{K}f\operatorname{dm} : K\subset\Omega, K\text{ is compact}\right\}?$$

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What do you mean by limited over each compact $K$? Bounded? Does $C$ depend on $K$? –  Julian May 13 '13 at 11:48
    
$C$ does not depend on $K$. –  user34870 May 13 '13 at 12:12
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I think it follows at once from the monotone convergence theorem, and a little compactness argument to show that any compact $K$ is included in $K_j$ for large enough $j$. –  Olivier Bégassat May 13 '13 at 12:15

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Since for each $j$, we have $$\int_{K_j}f\mathrm dm\leqslant\sup_{\substack{K\subset \Omega,\\ K\mbox{ compact}}}\int_Kf\mathrm dm=:M,$$ we get $\leqslant $ taking the limit $j\to \infty$.

For the reverse inequality, take $\varepsilon\gt 0$ and $K\subset \Omega$ a compact set such that $M-\varepsilon\lt\int_Kf\mathrm dm$. Since $K\subset \bigcup_{j\geqslant 1}\operatorname{int}(K_j)$, we have for some $J$ that $K\subset\bigcup_{j=1}^J\operatorname{int}(K_j)\subset \bigcup_{j=1}^JK_j=K_J$, hence $$M-\varepsilon\lt\int_Kf\mathrm dm\leqslant \int_{K_J}f\mathrm dm\leqslant\lim_{j\to \infty}\int_{K_j}f\mathrm dm.$$ As $\varepsilon$ was arbitrary, we conclude $M\leqslant \lim_{j\to \infty}\int_{K_j}f\mathrm dm$.

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