Sigma Algebras and Measure

Question

Let $(X,A,\mu )$ be a measure space (where denotes the measure and $A$ the $\sigma$-algebra) and $u : X \to \overline{\Bbb R}$ (where $\overline{\Bbb R}$ includes $-\infty, +\infty$) be a non-negative $A$-measurable function with $\int_X ud\mu < \infty$ (i.e $u$ is $\mu$-summable).

Explain why the set $\{x \in X : u(x) = +\infty\}$ belongs to the $\sigma$-algebra $A$ and prove that $\mu\{x \in X : u(x) = +\infty\} = 0$

I am not quite sure how to approach this. I know the properties of a $\sigma$-algebra but I do not know how to explain that something is in a sigma algebra. I suppose for the second part about the measure of the set, I have to use Fatous lemma since it is specified that $u$ is non-negative and $\mu$-summable but I am not sure about that either. Would I have to use dominated convergence perhaps?

Answer 1

For $n \in \mathbb N$ let $M_n:=\{x \in X: u(x)>n\}.$ Then $M_n \in A$, since $u$ is $A$- measurable, and

$$\{x \in X : u(x) = +\infty\} = \bigcap_{n \in \mathbb N}M_n.$$

Can you proceed ?

Answer 2

The fact that the set belongs to $\mathcal A$ is part of the definition of measurability of extended real valued functions. There is nothing to prove! Now suppose $\mu \{x:u(x)=\infty\}>0$. Then $\int u d\mu \geq \int _{\{x:u(x)=\infty\}}d \mu =(\infty) (\mu \{x:u(x))=\infty\}=\infty$.

Answer 3

In your question it is mentioned that $u:X\to\overline{\mathbb R}$ is $\mathcal A$-measurable.

Then the question rises: with respect to which $\sigma$-algebra on $\overline{\mathbb R}$?

If it is a $\sigma$-algebra with the property that singleton subset $\{+\infty\}$ is measurable (and usually it is) then it is actually immediate that $\{x\in X\mid u(x)=+\infty\}$ is measurable. This simply because $u$ is measurable, so that $u^{-1}(\{+\infty\})\in\mathcal A$.

If it is known that the $\sigma$-algebra on $\overline{\mathbb R}$ contains sets like $(x,+\infty]$ where $x\in\mathbb R$ then it can be shown that also $\{+\infty\}$ is an element of the $\sigma$-algebra. For that see the proof of Fred.

For the question concerning $\mu(\{x\in X\mid u(x)=+\infty\})=0$ see the answer of Kavi.