I am trying to prove that given two random variables $X,Y$, $F_X=F_Y$ implies $P_X=P_Y$.

I was given a hint to first prove the following corollary:

let $\mathcal{F}$ be a sigma-algebra generated by an algebra $\mathcal{F}_0$, and let $P_1,P_2$ probability functions such that $P_1=P_2$ over $\mathcal{F}_0$. Prove that $P_1=P_2$.

My strategy is to define $A:=\{E\in\mathcal{F} \ | \ P_1(E)=P_2(E)\}$ and show that A is a sigma-algebra. And since $\mathcal{F}_0\subseteq A $ and $\mathcal{F}=\sigma(\mathcal{F}_0)$ we get $\mathcal{F}\subseteq A$.

I just haven't managed to prove that $A$ is closed under countable union. I would like to say that given $\{ E_i\}_{i\in I}\subseteq A$:

$ P_1(\bigcup E_{i})=P_1(\sqcup E'_{i})=\sum{P1(E'_i)}=\sum{P_2(E'_i)} = \ ... = P_2(\bigcup E_{i}) $

With $E'_i$ being pairwise disjoint ($E'_n:=E_n \setminus E'_{n-1}$)

but obviously I cannot write the third equality like that because I can't assure that the sets $E'_i$ are in fact in $A$.

How can I prove that a union/interection/difference of two sets in $A$ is in $A$?


1 Answer 1


I think you can prove the hint using Dynkin's $\pi$-$\lambda$ Theorem which states that given $P$, a $\pi$-system, and $L$, a $\lambda$-system, if $P \subseteq L$, then $\sigma(P) \subseteq L$.

In the context of the hint you described, $\mathcal{F}_0$ is an algebra and is, in turn, a $\pi$-system, thus all that remains is to check that the set $A$ you defined is a $\lambda$-system and contains $\mathcal{F}_0$ (but the containment is clear by the definition of $A$)

There are three criteria to check for $A$ to be a $\lambda$-system;

$(1)$: $\Omega \in A$, but $\Omega \in \mathcal{F}$ and $P_1(\Omega) = 1 = P_2(\Omega)$.

$(2)$: If $B, C \in A$, $B \subseteq C$ then $C \setminus B \in A$, but $P_1(C\setminus B) = P_1(C) - P_1(B) = P_2(C) - P_2(B) = P_2(C \setminus B)$. Lastly,

$(3)$: If $B_n \in A$ for all $n$, and $B_n \uparrow B$ then $B \in A$, but this follows from continuity of measures, namely,

$$P_1(B) = \lim_{n \to \infty}P_1(B_n) = \lim_{n \to \infty} P_2(B_n) = P_2(B).$$

I think this is enough, because we see that $A$ is a $\lambda$-system containing $\mathcal{F}_0$ and thus application of the $\pi$-$\lambda$ Theorem gives us that $\sigma(\mathcal{F}_0) \subseteq A$.

  • $\begingroup$ It does seem to solve my problem, but there should be a more elementary solution, since our course hasn't covered this material yet. $\endgroup$
    – Guy
    Commented Nov 7, 2015 at 13:33

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