Let $(X, \mathcal X, \mu)$ be a $\sigma$-finite measure space and $f, g:X \to \mathbb R \cup \{\pm \infty\}$ $\mu$-measurable.
- We say that $\int f \mathrm d \mu$ exists if at least $\int f^+ \mathrm d \mu < +\infty$ or $\int f^- \mathrm d \mu < +\infty$. In this case, $$ \int f \mathrm d \mu := \int f^+ \mathrm d \mu - \int f^- \mathrm d \mu \in \mathbb R \cup \{\pm \infty\}. $$
- We say that $\int f \mathrm d \mu$ is finite if $\int f^+ \mathrm d \mu < +\infty$ and $\int f^- \mathrm d \mu < +\infty$. In this case, $\int f \mathrm d \mu \in \mathbb R$.
We adopt the convention that $a + (\pm\infty) = (\pm\infty)$ for all $a \in \mathbb R$. I'm trying to prove the following statement, i.e.,
Theorem If $\int f \mathrm d \mu$ exists and $\int g \mathrm d \mu$ is finite, then $\int (f+g) \mathrm d \mu$ exists and $$ \int (f+g) \mathrm d \mu = \int f \mathrm d \mu + \int g \mathrm d \mu. $$
In below attempt, I'm stuck at proving $\int (f+g)^- \mathrm d \mu = +\infty$. Could you elaborate on my difficulty?
Proof: WLOG, we assume $\int f^+ \mathrm d \mu < \infty$. We have $(f+g)^+ \le f^+ + g^+$, so $$ \int (f+g)^+ \mathrm d \mu \le \int (f^+ + g^+) \mathrm d \mu = \int f^+ \mathrm d \mu + \int g^+ \mathrm d \mu < +\infty. $$
As such, $\int (f+g) \mathrm d \mu$ exists. If $\int f^- \mathrm d \mu <+ \infty$, the claim follows from the linearity of integral. Now we consider the case $\int f^- \mathrm d \mu =+ \infty$. Then $\int f \mathrm d \mu =- \infty$. It suffices to prove that $\int (f+g)^- \mathrm d \mu = +\infty$.
