I have been asked to prove that if probability measures $P$ and $Q$ defined on $(\mathbb{R}, \mathcal{B})$, where $\mathcal{B}$, is the borel sigma algebra, then the set $\mathcal{F} = \{A \in \mathcal{B}: P(A) = Q(A)\}$ is a sigma algebra.
Prove that the collection of sets where two finite measures agree in not necessarily a sigma-algebra. proves that this is not necessarilly true for general finite measures and general measure spaces. So I guess if its true the proof should use properties of probabilities and borel sets. We have to show that (1) the empty set is in $\mathcal{F}$, (2) that if $A \in \mathcal{F}$ then so is the complement: $A^c \in \mathcal{F}$, and that (3) a countable union of sets in $\mathcal{F}$ is in $\mathcal{F}$.
(1) Follows since P and Q must be zero on the empty set.
(2) Follows since if $A \in \mathcal{F}$ then $P(A) = Q(A)$ so $1-P(A^c) = 1 – Q(A^c)$ so $P(A^c) = Q(A^c)$.
(3) I am stuck in this one, suppose $A_i \in \mathcal{F}$ for $i = 1,2,…$. I have tried defining a sequence $B_1,B_2,…$ such that $\cup A_i = \cup B_i$ and the $B_i$'s are disjoint ($B_1 = A_1$ and $B_n = A_n \setminus \cup_{i=1}^{n-1} A_i$) so that the probability of $\cup A_i$ can be written as the sum of the probabilities of $B_i$'s however I dont see anyway to establish that $B_i$'s belong in $\mathcal{F}$, for example I dont see that P and Q should coincide in $A_2 \setminus A_1$. Also Im not able even for a simple example like $A_1 \cup A_2$ since $P(A_1 \cup A_2) = P(A_1) + P(A_2) – P(A_1 \cap A_2)$ and I dont see why $P(A_1 \cap A_2)$ and $Q(A_1 \cap A_2)$ could be the same (unless $A_1$ and $A_2$ are disjoint which is not necessarilly the case).
My guess is that im failing to use some important property of Borel sets or that the result does not hold, however I havent been able to come with a counterexample either.
Best Answer
The statement is false.
Counterexample:
Let $$P\{1\} = P\{2\} = P\{3\} = \frac{1}{3} = Q\{4\}\,,\quad Q\{1\} = \frac{2}{3}\,.$$
Then $P$ and $Q$ agree on $\{1, 2\}, \{1, 3\}$ but don't agree on their intersection. Thus, family of all sets on which $P$ and $Q$ agree cannot be $\sigma$-algebra.