I am trying to prove an analogous statement of the Lebesgue Monotone Convergence theorem on function domains.
Let $A \subset \mathbb{R}^n$ be a bounded set that has volume. Let $f : A \mapsto \mathbb{R}, f \ge 0$ be an unbounded function. Suppose $C_i$ is a sequence of compact sets with volume, $C_i \subset A$ with $C_i$ increasing to $A$. Then $f$ is integrable iff $f$ is integrable on each $C_i, \lim_{i \to \infty} \int_{C_i} f$ exists, and $\int_A f = \lim_{i \to \infty} \int_{C_i} f$.
This is intuitively clear. If $f$ is integrable on each $C_i$ and $\lim_{i \to \infty} \int_{C_i} f = I$, then I should be able to find $N_1, N_2$ such that $vol(A)-vol(C_i) < \epsilon$ if $i > N_1$ and $|\int_{C_i} f – I| < \epsilon$ if $i > N_2$ for some $\epsilon > 0$. Then take $N = max\{N_1,N_2\}$, I should be able to show $f$ is integrable and $|\int_A f – I| < \epsilon$. But I got stuck in trying to work out the details of partitions.
Conversely, if $f$ is integrable, I should be able to apply Darboux's theorem and show that $|\int_{C_i} f – \int_A f| < \epsilon$. But I think Darboux's theorem only applies to bounded functions, and I don't know that how to use the condition that each $C_i$ is compact. It would be nice if $f$ is continuous, then $f$ would be bounded on each $C_i$, but the proposition does not mention this either. Can someone help with these details or give me a hint? Thanks a lot!
Best Answer
Let $f_i=f\chi_{C_i}$. Then $(f_i)$ is a sequence of non-negative measurable functions increasing to $f$. By Lebesgue's Monotone convergence Theorem $\int_A f=\lim \int_A f_i=\lim \int_{C_I} f$ (and this does not require any further hypothesis like integrability, boundedness etc). The conclusion follows immediately from this.