# If $\mu(\omega, A)$ is a regular conditional distribution of $X$ given $\mathcal{F}$, explain the notation $\mu(\omega, dx)$

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).
• 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... – JacobsonRadical Nov 21 '19 at 13:37
• 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 $\int_{A}d\mu(\omega, x)$? – JacobsonRadical Nov 21 '19 at 13:37
• $\int_A \mu (\omega,dx)=\int I_A (x) \mu (\omega,dx)=\mu(\omega,A)$. – Kavi Rama Murthy Nov 21 '19 at 14:01