Consider a measure space $(\Omega, \mathcal{F}, \mu)$ and two integrable, measurable functions
$$f,g: \Omega \to [-\infty, + \infty]$$
I.e., $$\int fd\mu, \int g d \mu \in \mathbb{R}$$
I proved that in this case $$\mu\{f= \pm \infty\} = 0 = \mu\{g= \pm \infty\}$$
My book then gives the following remark:
For every $\alpha, \beta \in \mathbb{R}$, the function $\alpha f + \beta g$ is defined $\mu$-almost everywhere (the function is not defined when we encounter something of the form $+ \infty – (+ \infty), -\infty + (+ \infty))$
We formally define this expression to be $0$ in that case.
My book then proves that $\alpha f + \beta g$ is integrable then and that $$\int (\alpha f + \beta g) d \mu = \alpha \int f d\mu + \beta \int g d\mu$$
Question: How can I show that the function $\alpha f + \beta g$ is $\mathcal{F}$-measurable? It makes intuitive sense, because we 'ruin' the measurability in a set of measure $0$.
Best Answer
Essentially we need to show that the sum $f+g$ of measurable functions (with additional agreement $\pm\infty+\mp\infty=0$) is measurable. The proof is a slight modification of the standard proof for finite-valued functions. Namely, let $F_{\pm\infty}=\{x : f(x) = \pm\infty\}$ and similarly for $g$. Then for $a\in\mathbb{R}$ $$\{x : f(x) + g(x) < a\} = I(a) \cup \bigcup_{r\in\mathbb{Q}}\big(\{x : f(x) < r\} \cap \{x : g(x) < a - r\}\big),$$ where $I(a) = (F_{+\infty}\cap G_{-\infty})\cup(F_{-\infty}\cap G_{+\infty})$ if $a > 0$ and $I(a) = \varnothing$ otherwise. All the sets on the right-hand side are measurable, thus the one on the left-hand side is measurable too.