2
$\begingroup$

Let $\left(X,M,\mu\right)$ be a measurable space, $f: X \rightarrow \left[0,\infty\right] $ and $$\omega_{f}=\left\{\left(x,t\right)\in X\times \left[0,\infty\right]\::\:0\leq t\leq f(x)\right\}$$ Show that:

  • $f$ is measurable if and only if $\omega _{f} \in M \otimes \mathcal{B}_{\left[0,\infty\right]}$ where $\mathcal{B}_{\left[0,\infty\right]}$ is the Borel $\sigma$-algebra and $M \otimes \mathcal{B}_{\left[0,\infty\right]}$ is $\sigma$-algebra generated by $M \times \mathcal{B}_{\left[0,\infty\right]}$.
  • If $f$ is measurable, then $\int_{X}fd\mu=\Pi\left(\omega_{f}\right)$ where $\Pi$ is the measure in $M \otimes \mathcal{B}_{\left[0,\infty\right]}$ such that $\Pi \left(A\times B\right)=\mu \left(A\right) \lambda \left(B\right)$ where $\lambda$ is a measure in $\mathcal{B}_{\left[0,\infty\right]}$ ($\lambda$ can be Lebesgue Measure).

Remark: I have tried the following:

Note that: $$\omega _{f}=\left(X \times \left[0,\infty\right] \right) \cap \left\{\left(x,t\right)\in X\times \left[0,\infty\right] \: : \: f(x)-t \geq 0 \right\}.$$ We know that $ X\times \left[0,\infty\right] \in M \otimes \mathcal{B}_{\left[0,\infty\right]}$ and $\left\{\left(x,t\right)\in X\times \left[0,\infty\right] \: : \: f(x)-t \geq 0 \right\} \in M \otimes \mathcal{B}_{\mathbb{R}}$, but we need $\left\{\left(x,t\right)\in X\times \left[0,\infty\right] \: : \: f(x)-t \geq 0 \right\} \in M \otimes \mathcal{B}_{\left[0,\infty\right]}$, precisely that is what I could not prove.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.