Question. Let $U \subset \mathbb{R}^m$ be an open $J$-measurable set and $(K_i)$ a sequence of compact $J$-measurable sets such that $K_i \subset \operatorname{int} K_{i+1}$ for every $i \in \mathbb{N}$ and $U = \bigcup_{i} K_i$. Prove that, for every integrable function $f: U \rightarrow \mathbb{R}$, we have$$\int_{U} f(x) dx = \lim_{i \rightarrow \infty} \int_{K_i} f(x) dx.$$
Initially, I thought it would be sufficient to observe that$$\left|\int_{U} f(x) \, dx - \int_{K_i} f(x) \, dx \right| = \left|\int_{U-K_i} f(x) \, dx\right| \leq M \operatorname{vol}(U-K_i),$$where $M$ denotes the maximum value of $f$ on $U$ (assuming $f$ is bounded, even though the exercise does not explicitly state this). We can see that as $\operatorname{vol}(U-K_i)$ approaches zero, the result would follow. However, this argument does not seem to take into account the compactness of $K_i$, the openness of $U$, or the condition $K_i \subset \operatorname{int}K_{i+1}$. How should I proceed?
EDIT: We also have a convergence result that allows us the pass the limit inside the integral when the sequence is uniformly convergent, so another approach would be to define $f_i = f\,\chi_{K_i}$ and integrate over $U$ for example. However, we would need to show that $f_i$ are integrable and that the convergence is uniform, which I'm not certain whether it is.