If we have a probability space $(\Omega,\mathcal{F},P)$ and $\Omega$ is partitioned into pairwise disjoint subsets $A_{i}$, with $i\in\mathbb{N}$, then the law of total probability says that $P(B)=\sum_{i=1}^{n}P(B|A_{i})P(A_i{})$. This law can be proved using the following two facts:
\begin{align*}
P(B|A_{i})&=\frac{P(B\cap A_{i})}{P(A_{i})}\\
P\left(\bigcup_{i\in \mathbb{N}} S_{i}\right)&=\sum_{i\in\mathbb{N}}P(S_{i})
\end{align*}
Where the $S_{i}$'s are a pairwise disjoint and a $\textit{countable}$ family of events in $\mathcal{F}$.
However, if we want to apply the law of total probability on a continuous random variable $X$ with density $f$, we have (like here):
$$P(A)=\int_{-\infty}^{\infty}P(A|X=x)f(x)dx$$
which is the law of total probabillity but with the summation replaced with an integral, and $P(A_{i})$ replaced with $f(x)dx$. The problem is that we are conditioning on an $\textit{uncountable}$ family. Is there any proof of this statement (if true)?
Best Answer
Think of it like this: Suppose you have a continuous random variable $X$ with pdf $f(x)$. Then $P(A)=E(1_{A})=E[E(1_{A}|X)]=\int E(1_{A}|X=x)f(x)dx=\int P(A|X=x)f(x)dx$.