# Monotone convergence theorem for increasing negative functions

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.)

• I'm not sure if this answers your question, but you can just apply the original MCT to $g_k-g_1$ to obtain the result. May 24, 2021 at 18:30

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.