The convergence of the integral is well-defined

Question

Let's begin with a definition:

Let $U\subset \mathbb R^m$ be an open set and $K_i$ be a sequence of compact, J-measurable sets such that $U=\cup K_i$ and $K_i\subset\operatorname{int}K_{i+1}$ for every $i\in \mathbb N$. Given a continuous function $f:U\to \mathbb R$, we say the integral $\int_Uf(x)dx$ is convergent when for every such sequence $\{K_i\}$, there exists $\lim_{i\to \infty}\int_{K_i}f(x)dx$.

I want to prove if we find only one such sequence, then the integral of a non-negative function turns out to be convergent.

Any ideas how to begin?

Answer 1

Let $f\ge0$ and $\{C_i\}$ be a sequence of compact sets exhausting $U$ such that $\int_{C_i}f$ converges to, say, $I$. If $\{K_i\}$ is any other sequence of compact sets exhausting $U$, show that $\int_{K_i}f$ is increasing and bounded by $I$.