# Can we prove the law of total probability for continuous distributions?

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 distribution $f$, we have (like here): $$P(A)=\int_{\Omega}P(A|x)f(x)dx$$ 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)?

• Fortunately it is true; otherwise we would also have trouble with conditional expectation for continuous variables. I was not able to find the proof in my measure theory book this morning, skimming through it, but going the long way, you may be convinced by the proof of the law of total expectation together with the relation that $P(A\mid B) = E[1_A\mid B]$. – Therkel Nov 29 '16 at 7:54

Think of it like this: Suppose you have a continuous random variable $$X$$ with pdf $$f(x)$$. Then $$P(A)=E(1_{A})=E[E(1_{A}|X)]=\int E(1_{A}|X=x)f(x)dx=\int P(A|X=x)f(x)dx$$.
• The event $\{X = x \}$ has 0 probability for a continuous random variable $X$, so your $P(A | X = x) = P(A, X=x)/P(X = x)$ is not well defined. You can condition on any event $B \in \sigma(X)$ i.e. on any event from the sigma algebra generated by $X$. – baibo Feb 11 at 21:33