This is exercise 4.1.13 of Durrett $3^{rd}$, stating that

Let $\mu(\omega, A)$ be a regular conditional distribution of $X$ given $\mathcal{F}$, and let $f:(S, \mathcal{S})\longrightarrow(\mathbb{R},\mathcal{R})$ have $\mathbb{E}|f(X)|<\infty$. Start with simple functions and show that $$\mathbb{E}(f(X)|\mathcal{F})=\int \mu(\omega, dx)f(x)\ \text{a.s.}$$

Well, I got stuck at the first place, what is $\mu(\omega, dx)$? should the second coordinate of $\mu$ be a set $A\in\mathcal{S}$? and what is the meaning of the integral RHS?

I understand to show above equality, I need to first show that it is true for indicator function, and then extends to simple function using linearity and then using convergence theorem to extend the result to non-negative functions, and then general integrable function follows immediately by separating them to positive and negative parts.

However, for instance, to show this is true for indictor function $f=\mathbb{1}_{A}$ for $A\in\mathcal{S}$, since $\mu$ is already a regular conditional distribution, we have $$\mathbb{P}(X\in A|\mathcal{F})=\mathbb{E}(\mathbb{1}_{A}(X)|\mathcal{F})=\mu(\omega, A),$$ so this exercise actually implies $$\mu(\omega, A)=\int \mu(\omega, dx)f(x)????,$$

I am really confused by this exercise now.. could someone please explain me what is going on here?

Thank you so much!


$\int f(x)d\mu(x)$ is just another notation for $\int f d\mu$. This notation is useful when there are several variables involved. So $\int f(x)d\mu (\omega,x)$ is the integral of $f$ w.r.t. the measure $\mu_{\omega}$ defined by $\mu_{\omega}(E)=\mu (\omega, E)$ (with $\omega$ fixed).

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  • $\begingroup$ could you apply your answer to show the equality holds for $f=\mathbb{1}_{A}$? I tried to apply your argument to prove this exercise but cannot proceed... $\endgroup$ – JacobsonRadical Nov 21 '19 at 13:37
  • $\begingroup$ For instance, in this case, $\mathbb{E}(\mathbb{1}_{A}(X)|\mathcal{F})=\mathbb{P}(X\in A|\mathcal{F})=\mu(\omega, A)$, but then how could I express $\mu(\omega, A)$? is it $\int_{A}d\mu(\omega, x)$? $\endgroup$ – JacobsonRadical Nov 21 '19 at 13:37
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    $\begingroup$ $\int_A \mu (\omega,dx)=\int I_A (x) \mu (\omega,dx)=\mu(\omega,A)$. $\endgroup$ – Kavi Rama Murthy Nov 21 '19 at 14:01
  • $\begingroup$ Thank you so much! $\endgroup$ – JacobsonRadical Nov 21 '19 at 14:20

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