# Show that $\alpha f + \beta g$ is measurable when $f,g$ are measurable.

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 $$0$$.

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.
• I think your answer is wrong. The left set isn't contained in the right set. To see this,take $a >0$, $x$ s.t. $f(x) = \infty$ and $g(x) = -\infty$. Then there is no rational $r$ such that $f(x) < r$. – user370967 Oct 21 '18 at 20:35