On integration, measurability, almost everywhere concept

[kalish1]

Suppose $\int f d\mu < \infty.$ Let $$h(\omega)=\begin{cases}f(\omega) \ \ \ \text{if} \ \ f(\omega)\in \mathbb{R} \\ \\ 0 \ \ \ \text{if} \ \ \ f(\omega)=\infty\end{cases}$$
How to show $h$ is measurable and $\int f d\mu = \int h d\mu?$

**Attempt:** It is known that the product of two measurable functions is again measurable, and note that $h(\omega)=f(\omega)I_\mathbb{R}(f(\omega))$ *(is this formulation right?)*.

**Claim:** $I_\mathbb{R}(f(\omega))$ is measurable.

*Proof:* $\mathbb{R}$ is measurable because it is a Borel set. So for any $\alpha \in \mathbb{R},$

$$\left\{{f(\omega) \in {\mathbb{R}}: I_\mathbb{R} \left({f(\omega)}\right) \ge \alpha}\right\} = \begin{cases}\varnothing & \text{if $1 < \alpha$}\\
\mathbb{R} & \text{if $0 < \alpha \le 1$}\\
{\mathbb{R}} & \text{if $\alpha \le 0$}\end{cases}.$$

Since $\varnothing$ and $\mathbb{R}$ are measurable, we conclude that $I_\mathbb{R}(f(\omega))$ is measurable.

Now how can I show that the two integrals are equal?

And is my work above correct?

I have crossposted this question on real analysis - On integration, measurability, almost everywhere concept - Mathematics Stack Exchange

 

[Fallen Angel]

Hi kalish,

Just note that \(\displaystyle f\) having a finite integral implies \(\displaystyle \mu(\{\omega \in \mathbb{R} \ : \ f(\omega)=\infty\})=0\) , hence \(\displaystyle f=h\) a.e.p.