Here I have an exercise of book: Probability and Measure of PATRICK BILLINGSLEY of conditional probability in the page 442, exercice 33.4 (b):
Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, and a second probability space $(\mathbb{R}²,\mathscr{B}(\mathbb{R}^2),\mu)$ where $\mathscr{B}(\mathbb{R}^2)$ is the borel sigma algebra and $\mu$ the distribution of a randon vector $(X,Y)$ in $(\Omega, \mathcal{F}, \mathbb{P})$. Suppose that $(X, Y)$ has density $f$, and show that $$ \mathbb{P}[Y\in F|X]_{\omega}=\dfrac{\int_F f(X(\omega),t)dt}{\int_{} f(X(\omega),t)dt} $$ with probability $1$.
My problem: In the page $432$ in the example $33.5$ a similar problem is analyzed, but this example is easy because the sigma algebra in question is the sigma algebra generated by vertical strips $E\times \mathbb{R}$, which enables the use of Fubinni Theorem in order to get an iterated integral.In this case I do not know how to proceed.
For brevity of notation, set
$$g(x) := \frac{\int_F f(x,y) \, dy}{\int f(x,y) \, dy}. \tag{1}$$
To prove $$\mathbb{P}(Y \in F \mid X) = g(X)$$
it suffices to show $$\int_A g(X) \, d\mathbb{P} = \int_A \mathbb{P}(Y \in F \mid X) \, d\mathbb{P}$$ for any $A \in \sigma(X)$. Recall that any $A \in \sigma(X)$ can be written as $A= X^{-1}(B)$ for some Borel set $B$. Hence, by Tonelli's theorem,
$$\begin{align*} \int_A g(X) \, d\mathbb{P} &= \int_{\mathbb{R}^2} 1_B(x) g(x) \, d\mu(x,y) \\ &= \int_{\mathbb{R}} \int_{\mathbb{R}} 1_B(x) \, g(x) \, f(x,y) \, dx \, dy \\ &= \int 1_B(x) g(x) \left( \int f(x,y) \, dy \right) \, dx \\ &\stackrel{(1)}{=} \int 1_B(x) \left( \int_F f(x,y) \, dy \right) \, dx. \tag{2} \end{align*}$$
On the other hand,
$$\begin{align*} \int_A \mathbb{P}(Y \in F \mid X) \, d\mathbb{P} &= \int_A 1_F(Y) \, d\mathbb{P} = \int_{\mathbb{R}^2} 1_B(x) 1_F(y) \, d\mu(x,y) \\ &= \int \int 1_B(x) 1_F(y) f(x,y) \, dx \, dy. \tag{3} \end{align*}$$
Since $(2)$ equals $(3)$, we are done.