I don't quite understand the proof for this theorem: Let $(X,\Sigma)$ be a measurable space and let $E\in\Sigma$. If $f:X\to\mathbb{R}$ is $\Sigma$-measurable on $E$, then so is $|f|^{\beta}$ where $\beta\geq 0$.
Proof: Let $E=\{x\in X\mid\ f(x)\in\mathbb{R}\}$. Then $f(x)\in\{-\infty,\infty\}$ for all $x\in X\setminus E$, so $|f|^{\beta}$ is constant on $X\setminus E$, so it is $\Sigma$-measurable on $X\setminus E$.
Also $f(x)\in\mathbb{R}$ for all $x\in E$, so $\{x\in X\mid |f(x)|^{\beta}>\alpha\} = E$ (if $\alpha<0$) or $=\{x\in E\mid f(x)>\alpha^{1/\beta}\}\cup\{x\in E\mid f(x)<-\alpha^{1/\beta}\}$ (if $\alpha\geq 0$).
Hence $|f|^{\beta}$ is $\Sigma$-measurable on $E\cup(X\setminus E) = X$. So it is $\Sigma$-measurable on every $E'\in\Sigma$, in particular $E$.
What I don't understand is that we have chosen a specific $E\in\Sigma$, which seems not enough to prove the general case for any $E\in\Sigma$. Also $\Sigma$ was chosen such that $E\in\Sigma$, which is not true in general ($\Sigma$ could just be $\{X,\emptyset\}$ for example). What is it that I'm missing?
I think the following is a correct proof:
Given $E\in\Sigma$ and $\beta\geq 0$, and suppose $f:X\to\mathbb{R}$ is $\Sigma$-measurable on $E$. Then for all $\alpha\in\mathbb{R}$, $\{x\in E\mid\alpha<f(x)\}\in\Sigma$.
Now there are two cases to consider: If $\alpha\geq 0$, then $\{x\in E\mid\alpha<|f(x)|^{\beta}\}=\{x\in E\mid f(x)>\alpha^{1/\beta}\}\cup\{x\in E\mid f(x)<-\alpha^{1/\beta}\}$ which is a member of $\Sigma$. Conversely if $\alpha<0$, then $\{x\in E\mid\alpha<|f(x)|^{\beta}\}=E\in\Sigma$.
Hence in for all $\alpha\in\mathbb{R}$, $\{x\in E\mid\alpha<|f(x)|^{\beta}\}\in\Sigma$. So $|f|^{\beta}$ is $\Sigma$-measurable on $E$.