# Role of Radon-Nikodym theorem in definition of conditional probabilities

Let $(\Omega,\mathscr{F},P)$ be a probability space and $\mathscr{G}\subset\mathscr{F}$ a $\sigma$-field contained in $\mathscr{F}$. If we define the finite measure v on $\mathscr{G}$ by

$\text{v}(G)=P(A\cap G)$ for all $G\in\mathscr{G}$,

then my question concerns the role of the Radon-Nikodym theorem in the definition of conditional probabilities used by Patrick Billingsley (1995) in his textbook "Probability and Measure", which he denotes by $P[A\|\mathscr{G}]$, and defines by;

i) $P[A\|\mathscr{G}]$ being measurable $\mathscr{G}$,

ii) $\int_G P[A\|\mathscr{G}] \, dP=P(A\cap G):=\operatorname{v}(A\cap G)$ for all $G\in\mathscr{G}$.

If $P_0$ is $P$ restricted to $\mathscr{G}$, then since $\operatorname{v}\ll P_0$ for $P_0<\infty$, and therefore $\operatorname{v}<\infty$, I think I can see that the Radon-Nikodym theorem guarantees the existence of a real-valued, non-negative and $\mathscr{G}$-measurable function $f$, integrable w.r.t $P_0$ and hence $P$, satisfying i) and ii). Billingsley then labels $f:=P[A\|\mathscr{G}]$ (note: the definition above uses $P$ not $P_0$ I think since $\operatorname{v}(G)=\int_G f \, dP=\int_G f \, dP_0$ for all $G\in\mathscr{G}$).

My question is why does the Radon-Nikodym theorem guarantee that $f$ is a probability, since as far as I can tell it only guarantees it is non-negative? I can see that $f$ is a random variable with source probability space $(\Omega,\mathscr{F},P_0)$ and target space $(\mathbb{R},\mathscr{R},\mu)$ where $\mu$ is the distribution of $f$ satisfying $\mu(A)=P_0[f\in A]$ for all $A\in\mathscr{R}$, but that doesn't mean $f$ always lies in $[0,1]$ and so can be considered a probability?

• You don't need to write $a<<b$; you can write $a\ll b$. I edited accordingly. $\qquad$ – Michael Hardy Aug 2 '16 at 21:24

Consider the event $B_n:=\{f\ge 1+1/n\}$, where $n$ is a positive integer. Clearly $B_n\in\mathscr G$, and $$P_0(B_n)\ge P_0(A\cap B_n)=\int_{B_n}f\,dP_0\ge\int_{B_n}(1+1/n)\,dP_0=(1+1/n)P_0(B_n).$$ The only value of $P_0(B_n)$ consistent with this inequality is $P_0(B_n)=0$. Consequently $$P_0(f > 1)=P_0\left(\cup_{n=1}^\infty B_n\right)\le\sum_{n=1}^\infty P_0(B_n)=0.$$ This shows that $f\le 1$, $P_0$-a.s.
• Brilliant, thank you so much! I was going to try show this myself but I could never have come up with that. I wonder though why my textbook doesn't include something like this since it wasn't obvious to me. I sensed it doesn't follow immediately from the Radon-Nikodym theorem but depends on the measures involved - i.e. if $P_{0}$ is a general rather than a probability measure then $P_{0}(B_{n})=\infty$ might hold which means your proof wouldn't work then? Hmm.. maybe it is obvious now! – dandar Aug 3 '16 at 17:46