Let $f, g: X \to \overline{\mathbb{R}}$ be measurable (here $\overline{\mathbb{R}}$ denotes the extended real numbers).
Let $E_1 = \{ x \in X: f(x) = – \infty, g(x) = +\infty \}, E_2 = \{ x \in X: f(x) = + \infty, g(x) = -\infty \}$. Define
$$
h(x) =
\begin{cases}
f(x) + g(x) & \text{if $x \notin E_1 \cup E_2$} \\\\
0 & \text{if $x \in E_1 \cup E_2$}
\end{cases}
.$$
How can I show that $h$ is measurable? Preferably without using limits or such, but only with basic definitions like the definition of a measurable function, etc.
I am having a hard time because I don't know how to write the preimage $\{ x \in X: h(x) > a \}$ in terms of already-known measurable sets. I would appreciate any hints or answers.
For reference, this is from Bartle's Elements of Integration and Lebesgue Measure (I've paraphrased with some notation).
Best Answer
I'll suppress $x$ from the notation. Thus, $\{h>a\}:=\{x\in X:h(x)>a\}$.
First of all note that if $(q_n)_{n\in \mathbb{N}}$ is some enumeration of $\mathbb{Q}$, then $$ \{h>a, -\infty<f<\infty,-\infty<g<\infty\}=\bigcup_{n=1}^{\infty} \{q_n<f<\infty,a-q_n<g<\infty\} $$ Indeed, it's perfectly clear that the right-hand side is a subset of the left-hand side. On the other hand, if $x\in \{h>a, -\infty<f<\infty,-\infty<g<\infty\}, $ then $f(x)=h(x)-g(x)>a-g(x)$, implying, since $\mathbb{Q}$ is dense in $\mathbb{R}$ that there exists some $n_0\in \mathbb{N}$ such that $f(x)>q_{n_0}>a-g(x)$. This exactly means that $x\in \{q_{n_0}<f<\infty, a-q_{n_0}<g<\infty\}$, and since $x$ was arbitrary, we conclude that the left-hand side is a subset of the right-hand side and thus, the two sets must be the same. Now, since $f$ and $g$ are both measurable, the right-hand side is a union of measurable sets and we conclude that $\{h>a,-\infty<f<\infty,-\infty < g <\infty\}$ is measurable for any $a\in \mathbb{R}$.
To finish, note that if $a\geq 0$, then \begin{align} \{h>a,f=\infty\} &=\{f=\infty\}\setminus \{g=-\infty\}\\ \{h>a, g=\infty\} &= \{g=\infty\} \setminus \{f=-\infty\}, \end{align} both of which are measurable. Similarly, if $a<0$, then $\{h>a,f=\infty\}=\{f=\infty\}$ and $\{h>a,g=\infty\}=\{g=\infty\}$.
In conclusion, since $$ \{h>a\}=\{h>a, -\infty<f<\infty,-\infty<g<\infty\}\cup\{h>a,f=\infty\}\cup \{h>a,g=\infty\}, $$ all of which are measurable, we conclude that $\{h>a\}$ is measurable. Note that the difficulty here was really just finding a nice way to write the case where both functions were finite.