Let $(\Omega, \mathcal{F}, P)$ be a probability space, $\mathcal{G} \subseteq \mathcal{F}$ a $\sigma$-field in $\mathcal{F}$. Given $A \in \mathcal{F}$, the Radon-Nikodym theorem implies that there is ${ }^{[3]}$ a $\mathcal{G}$-measurable random variable $P(A \mid \mathcal{G}): \Omega \rightarrow \mathbb{R}$, called the conditional probability, such that $$ \int_{G} P(A \mid \mathcal{G})(\omega) d P(\omega)=P(A \cap G) $$ for every $G \in \mathcal{G}$, and such a random variable is uniquely defined up to sets of probability zero. A conditional probability is called regular if $\mathrm{P}(\cdot \mid \mathcal{B})(\omega)$ is a probability measure on $(\Omega, \mathcal{F})$ for all $\omega \in \Omega$ a.e.

I am used to the non-measure theoretic definition of conditional probability defined for events where $P(A|B)=\frac{P(A\cap B)}{P(B)}$. Why is this function not defined like

$$\frac{1}{P(G)}\int_{G} P(A \mid \mathcal{G})(\omega) d P(\omega)=\frac{P(A \cap G)}{P(G)} $$

which would make it in line with the definition for events? I understand that $P(G)=0$ is an issue, but what is the motivation of defining this representation for $P(A\cap G)$?

  • $\begingroup$ Note that the thing being defined is the random variable $P(A|\mathcal{G})$ and clearly, if a random variable fits into the one definition it fits into the other, except in the case where $P(G)=0$. $\endgroup$ Aug 7, 2022 at 22:33
  • $\begingroup$ What book is this from? $\endgroup$
    – littleO
    Aug 7, 2022 at 22:38
  • 1
    $\begingroup$ @littleO en.wikipedia.org/wiki/Conditional_probability_distribution But I was reading Billingsley Probability and Measure $\endgroup$ Aug 7, 2022 at 22:38

1 Answer 1


In the case where $\mathcal G$ is the $\sigma$-algebra generated by event $G$, $P(A \mid \mathcal G)$ is (up to sets of measure $0$) $P(A|G)$ when $G$ is true, $P(A|G^c)$ when $G$ is false. The equation then says $P(A \cap G) = P(G) P(A | G)$ and $P(A \cap G^c) = P(G^c) P(A | G^c)$, which fits with your non-measure theoretic definition.

  • $\begingroup$ Does this mean that $P(A|\mathcal{G})$ takes on exactly two values? And if I had events $B_1,\dots,B_n$ partitioning $\Omega$, then $P(A|\sigma(B_1,\dots,B_n))$ would take on $n$ values where $P(A|\sigma(B_1,\dots,B_n)) (\omega)=P(A|B_i)$? Is $P(A|\mathcal{G})$ doing the same type of "smoothing" that conditional expectation is doing? $\endgroup$ Aug 7, 2022 at 22:41
  • $\begingroup$ Yes it does (up to sets of measure $0$). $\endgroup$ Aug 7, 2022 at 23:17

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