5
$\begingroup$

Let $P$ and $Q$ two probability on $\mathbb R$ and let $$\mathcal A=\left\{\left[\frac{i}{2^n},\frac{i+1}{2^n}\right)\mid i\in \mathbb Z, n\in\mathbb N\right\}.$$

We have that $P(A)=Q(A)$ for all $A\in \mathcal A$. And I need to show that $P=Q$ on $\mathbb R$. I have already proved that $\mathcal B(\mathbb R)=\sigma (\mathcal A)$, and thus, I think I have to prove that $\mathcal A$ is a $\pi-$system. The problem, is a failed. First, I can take $A,B\in \mathcal A$ s.t. $A\cap B=\emptyset$, and $\emptyset\notin \mathcal A$, shall I add it in $\mathcal A$ ? And secondely, If $A\cap B\neq \emptyset$, I think that I have to show that either $A\subset B$ or $B\subset A$, but I also failed. How can I conclude ?

1 Answers 1

2

  1. Feel free to add $\emptyset$ to $\mathcal{A}$. Certainly, $P(\emptyset)=Q(\emptyset)$. It should also be clear that $\mathcal{A}$ is otherwise not closed under pairwise intersections.

  2. If $A\cap B\neq\emptyset$, it is indeed the case that $A\subseteq B$ or $B\subseteq A$. Now assume that $$\left[\frac{i}{2^n},\frac{i+1}{2^n}\right)\cap \left[\frac{j}{2^k},\frac{j+1}{2^k}\right)\neq\emptyset$$ with $k\geq n$. Then $$\left[\frac{i}{2^n},\frac{i+1}{2^n}\right)=\left[\frac{i\cdot 2^{k-n}}{2^k},\frac{(i+1)\cdot 2^{k-n}}{2^k}\right),$$ so you just have to show that $$\left[\frac{i\cdot 2^{k-n}}{2^k},\frac{(i+1)\cdot 2^{k-n}}{2^k}\right)\cap \left[\frac{j}{2^k},\frac{j+1}{2^k}\right)\neq\emptyset$$ implies $$\left[\frac{j}{2^k},\frac{j+1}{2^k}\right)\subseteq\left[\frac{i\cdot 2^{k-n}}{2^k},\frac{(i+1)\cdot 2^{k-n}}{2^k}\right)$$ and you are done.