TITLE: Question about Lebesgue integration on $\mathbb{R}^N$
QUESTION [3 upvotes]: 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 
$$\lim_{j\to\infty}\int_{K_j}f\operatorname{dm} =\sup\left\{\int_{K}f\operatorname{dm} : K\subset\Omega, K\text{ is compact}\right\}?$$

REPLY [1 votes]: 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 $M\leqslant \lim_{j\to \infty}\int_{K_j}f\mathrm dm$.
{"set_name": "stack_exchange", "score": 3, "question_id": 390315}