Let $\Omega \subset \mathbb{R}^n$ open and $f:\Omega \rightarrow [0, \infty[$ a measurable function. Suppose that there exist $C>0$ such that $$\int_K f dm < C,\ \forall\ K\subset\Omega,\ K\ \mbox{compact}.$$ Let $(K_j)$ be a sequence of subsets of $\Omega$ satisfying: $$K_j \subset \operatorname{int} (K_{j+1}), \ \bigcup_{n=1}^{\infty} K_j = \Omega .$$ I need to prove that
$$\lim_{j \rightarrow \infty} \int_{K_j} f dm = \sup \{ \int_K f dm: K \subset \Omega, K \mbox{compact} \} . $$
If we can prove that $m(\partial K)=0$, we can use that $\int_K f dm = \int_{\operatorname{int}(K)} f dm $ and find a sequence $(\int_{\operatorname{int}(K_j)} f dm)$ with $$\int_{\operatorname{int}(K_j)} f dm \rightarrow \sup\{ \int_{\operatorname{int}(K_j)} f dm : j \in \mathbb{N}\}.$$ Now we note that that $\{ \int_K f dm: K \subset \Omega, K \ \mbox{compact} \} \subset \{ \int_K f dm: K \subset \Omega, K\ \mbox{compact} \}$, and we can find an inequality. For another I thought to write $K = \bigcup_{n=1}^{\infty} K_{j} $ and try to play with their interior.
Do you think that I'm in the right away? And the question about boundary is true?
As commenters explained, the boundary of a compact set can have positive measure. How to proceed with the proof of $$\lim_{j \rightarrow \infty} \int_{K_j} f\, dm = \sup \left\{ \int_K\, f dm: K \subset \Omega, \ K \mbox{compact} \right\} $$ then?
The inequality $\le$ holds because every $K_j$ is also eligible for the supremum on the right.
To prove $\ge$, it suffices to show that for every compact $K$ there is $j$ such that $K\subset K_j$. Hint: since $\bigcup_j \operatorname{int}K_j=\Omega $, you have an open cover of $K$.