# Collection of borel sets where two probability measures coincide is a sigma algebra

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.

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.