MHB On integration, measurability, almost everywhere concept

kalish1
Messages
79
Reaction score
0
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
 
Physics news on Phys.org
Hi kalish,

Just note that $$f$$ having a finite integral implies $$\mu(\{\omega \in \mathbb{R} \ : \ f(\omega)=\infty\})=0$$ , hence $$f=h$$ a.e.p.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
Back
Top