Let $(X, \mathcal X, \mu)$ be a $\sigma$-finite measure space and $f, g:X \to \mathbb R \cup \{\pm \infty\}$ $\mu$-measurable, i.e., they are $\mu$-a.e. pointiwse limits of some sequences of $\mu$-simple functions.
- 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 $\int f \mathrm d \mu$ exists and $\int f \mathrm d \mu \in \mathbb R$.
We adopt the convention that for all $a \in \mathbb R$,
- $a + (+\infty) = (+\infty) + a = +\infty$,
- $a + (-\infty) = (-\infty) + a = -\infty$,
- $(-\infty) + (-\infty) = -\infty$, and
- $(+\infty) + (+\infty) = +\infty$.
Then we have
Lemma 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 order to justify Step 2 in the proof of Theorem 3.2 in the paper Existence and stability results in the $L^1$ theory of optimal transportation by Ambrosio/Pratelli, I come across below result, i.e.,
Theorem If $\int f^+ \mathrm d \mu < +\infty$ and $\int g^+ \mathrm d \mu < +\infty$, then $\int (f+g) \mathrm d \mu$ exists and $$ \int (f+g) \mathrm d \mu = \int f \mathrm d \mu + \int g \mathrm d \mu. $$
Could you have a check on my below attempt?
Proof Clearly, $\int f \mathrm d \mu$ and $\int g \mathrm d \mu$ exist. Moreover, $\{\int f \mathrm d \mu, \int g \mathrm d \mu\} \subset \mathbb R \cup \{-\infty\}$ and thus $\int f \mathrm d \mu+ \int g \mathrm d \mu \in \mathbb R \cup \{-\infty\}$. We have $(f+g)^+ \le f^+ + g^+$, so $$ \int (f+g)^+ \mathrm d \mu \le \int (f^+ + g^+) \mathrm d \mu \overset{(\star)}{=} \int f^+ \mathrm d \mu + \int g^+ \mathrm d \mu < +\infty. $$
Here $(\star)$ is becasue $f^+$ and $g^+$ are non-negative. As such, $\int (f+g) \mathrm d \mu$ exists. Moreover, $\int (f+g) \mathrm d \mu \in \mathbb R \cup \{-\infty\}$.
If $\int f^- \mathrm d \mu < +\infty$ then $\int f \mathrm d \mu$ is finite. The claim then follows from Lemma. Similarly, the claim holds if $\int g^- \mathrm d \mu < +\infty$.
Now we consider the case $\int f^- \mathrm d \mu = \int g^- \mathrm d \mu = +\infty$. In this case, $\int f \mathrm d \mu + \int g \mathrm d \mu = -\infty$. It suffices to prove $\int (f+g)^- \mathrm d \mu = +\infty$. We have $$ f^-=(-f)^+ = ((-f-g)+g)^+ \le (-f-g)^+ + g^+=(f+g)^-+g^+. $$ It follows that $$ \begin{align} \int f^- \mathrm d \mu &\le \int ((f+g)^- + g^+) \mathrm d \mu \\ &\overset{(\star\star)}{=} \int (f+g)^- \mathrm d \mu + \int g^+ \mathrm d \mu. \end{align} $$ Here $(\star\star)$ is because $(f+g)^-$ and $g^+$ are non-negative. Because $\int g^+ \mathrm d \mu <+\infty$, we can subtract it from both sides of $\int f^- \mathrm d \mu \le \int (f+g)^- \mathrm d \mu + \int g^+ \mathrm d \mu$. Then we get $$ \int (f+g)^- \mathrm d \mu \ge \int f^- \mathrm d \mu - \int g^+ \mathrm d \mu = +\infty -\int g^+ \mathrm d \mu = +\infty. $$
This completes the proof.
