This question is about the MCT for Lebesgue integration. Specifically, let $(X,S,\mu)$ be a measure space if $0 \le f_1 \le f_2 \le f_3 …$ is a sequence of increasing $S$-measurable functions from $X$ to $[0, \infty]$ and $f$ is defined to be the limit of $f_1, f_2, f_3, …$ then $\int f d\mu = lim_{k \rightarrow \infty} \int f_k d\mu$.
Assuming the above theorem, I want to prove that for $S$-measurable $g_1 \le g_2 \le g_3 …$ such that $g_k \le 0$ for all $k$ AND $\int |g_1| d\mu < \infty$, we also have $\int (lim_{k \rightarrow \infty} g_k) d\mu = lim_{k \rightarrow \infty} \int g_k d\mu$.
I am actually not sure if we can use the original MCT to prove this other version of the MCT. My idea is that we should add a constant $C$ to all the $g_k$'s so that the MCT applies to them then work backward to recover the desired result. This $C$ should be $\inf_{X} g_1$, but this could be $-\infty$; however, since $\int |g_1| d\mu < \infty$ we can (probably) find some other $C$ that works (we may have to restrict the integrals of the $g_k$'s to a marginally smaller subset of $X$ or something similar).
Could this approach work? I am not so sure because even if we could find such a $C$ that the MCT applies to the $g_k$'s then we still might not be able to recover the result we want because we might get $\infty – \infty$.
(I am using Sheldon Axler's Measure, Integration & Real Analysis.)
Best Answer
As pointed out in the comments, define $f_{k} := g_{k} - g_{1}$. Then $0 \le f_{1} \le f_{2} \le \cdots$. If $f := \lim f_{k} = \lim g_{k} - g_{1} = g-g_{1}$, by the original MCT: $$\int f d\mu = \lim \int f_{k} d\mu = \lim\int g_{k}d\mu - \int g_{1}d\mu $$ Hence, $$\int g d\mu - \int g_{1}d\mu = \lim\int g_{k}d\mu - \int g_{1}d\mu$$ Now, because $\int |g_{1}|d\mu <+\infty$, you can cancel $\int g_{1}d\mu$ on both sides and obtain the result.