Let $X$ be a non-negative integral random vairable on $(\Omega,\mathcal{F},\mathbb{P})$ and let $\mathcal{G}\subseteq\mathcal{F}$ be a sub-$\sigma$-algebra. I want to show that $$\mathbb{E}(X|\mathcal{G})=\int_{0}^{\infty}\mathbb{P}(X>t|\mathcal{G})dt.$$
I have done most part, but I had a hard to time to show that $$\int_{0}^{\infty}\mathbb{P}(X>t|\mathcal{G})dt$$ is $\mathcal{G}$-measurable.
To show the measurability, define $h(t,\omega):=\mathbb{E}(\mathbf{1}_{\{X>t\}}|\mathcal{G})(\omega)$, and my idea is to show that $h(t,\omega)$ is $(\mathcal{B}(\mathbb{R})\otimes\mathcal{G},\mathcal{B}(\mathbb{R}))$-measurable. Then, the desired measurability follows from Fubini-Tonelli theorem.
However, what I know is that $h(t,\omega)$ is $\mathcal{G}$-measurable for each $t$. Also, what I can say more is that since $X$ is a random variable, a regular conditional distribution always exists for $X$ with respect to $\mathcal{G}\subseteq\mathcal{F}$, and thus we can take $\mathbb{P}(X>t|\mathcal{G})$ to be a version of this regular conditional probability distribution.
But then what should I do? My idea is to show that the following collection $$\mathcal{E}:=\{(a,\infty):h^{-1}((a,\infty))\in \mathcal{B}(\mathbb{R})\otimes\mathcal{G}\}$$ is a $\sigma$-algebra. If we can show this, then since $\mathcal{E}$ contains a generating set of $\mathcal{B}(\mathbb{R})$, it follows that $\mathcal{E}=\mathcal{B}(\mathbb{R})$. It is not hard to show that it is closed under countable union and complement since the preimage of the function preserves these. However, how to show that $\mathcal{E}$ is not empty and contains the whole real line?
Am I heading to a correct direction?
Edit1: Attempt but not sure
Since $X$ is a random variable, a regular conditional distribution always exists for $X$ with respect to $\mathcal{G}\subseteq\mathcal{F}$, and thus we can take $\mathbb{P}(X>t|\mathcal{G})$ to be a version of this regular conditional probability distribution. In particular, $F(t,\omega):=1-h(t,\omega)$ is then a regular distribution function so that for each $\omega\in\Omega$, $F(t,\omega)$ is a distribution function on $\mathbb{R}$. Hence, $$h^{-1}(\mathbb{R})\in \{0\}\times A,\ \text{for any}\ A\in\mathcal{G}.$$ Since $\{0\}\in\mathcal{B}(\mathbb{R})$, it follows that $h^{-1}(\mathbb{R})$ belongs to the product $\sigma$-algebra.
Am I correct?
Best Answer
Let $\kappa : \Omega \times \mathcal{B}(\mathbb{R}) \to [0, 1]$ be a regular conditional probability kernel for $X$ given $\mathcal{G}$. We have $P(X > t \mid \mathcal{G})(\omega) = \kappa(\omega, \{X > t\}) = \int_{\mathbb{R}}1_{(t, \infty)}(x)\kappa(\omega,dx) = \int_{\mathbb{R}}f(t,x)\kappa(\omega,dx)$, where $f(t, x) = 1_{(t, \infty)}(x)$. Note that $f$ is measurable. Now let $$\mathcal{C} = \{g : \mathbb{R} \times \mathbb{R} \to [0, \infty) \mid (t, \omega) \mapsto \int_{\mathbb{R}}g(t, x)\kappa(\omega, dx)\text{ is $\mathcal{B}(\mathbb{R}) \otimes \mathcal{G}$-measurable }\}.$$ If $g(t, x) = 1_A(t)1_B(x)$ with $A, B \in \mathcal{B}(\mathbb{R})$, then $\int_{\mathbb{R}}g(t, x)\kappa(\omega, dx) = 1_A(t)\kappa(\omega, B)$ is $\mathcal{B}(\mathbb{R}) \otimes \mathcal{G}$-measurable. Then since the collection of sets whose indicator functions are in $\mathcal{C}$ can be shown to be a $\lambda$-system, by the $\pi-\lambda$ theorem, $\mathcal{C}$ contains all indicators of measurable sets. Then by linearity of integration and monotone convergence, $\mathcal{C}$ contains all $\mathcal{B}(\mathbb{R}) \otimes \mathcal{G}$-measurable functions.