Let $P_1$ and $P_2$ be two probability measures on a measurable space, $(\Omega, \mathcal{F})$. Then $P_1$ and $P_2$ are mutually singular (denoted $P_1 \perp P_2$) if there exists $A \in \mathcal{F}$ such that $P_1(A) = 1$ and $P_2(A) = 0$. The book Gaussian Random Processes by Ibragimov and Rozanov gives the following sufficient condition for mutual singularity of $P_1$ and $P_2$.
If there exists $A_1, A_2,.... \in \mathcal{F}$ such that $\lim\limits_{n\rightarrow\infty}P_1(A_n) = 0$ and $\lim\limits_{n\rightarrow\infty}P_2(A_n) = 1$, then $P_1 \perp P_2$.
They give the following explanation as to why, but I'm still unsure as to why this implies mutual singularity.
We can decompose $P_2$ as $P_2 = P_2^\prime + P_2^{\prime\prime}$ where $P_2^\prime \perp P_1$ and $P_2^{\prime\prime} << P_1$. Under the condition $\lim\limits_{n\rightarrow\infty}P_1(A_n) = 0$, we have, \begin{eqnarray}\lim\limits_{n\rightarrow\infty}P_2(A_n) &=& \lim\limits_{n\rightarrow\infty}P_2^\prime(A_n) + \lim\limits_{n\rightarrow\infty}P_2^{\prime\prime}(A_n)\\ &=& \lim\limits_{n\rightarrow\infty}P_2^\prime(A_n)\leq P_2^\prime(\Omega) < 1 \end{eqnarray}
I can follow the argument they give, but not the conclusion. Why can we deduce that $P_1 \perp P_2$ in this case? If we look at the definition of mutual singularity, what are the authors using as the measurable set $A$ here?
If $P_2'(\Omega) <1$ we get a contradiction to the assumption that $P_2(A_n) \to 1$. Hence $P_2'(\Omega) =1$. But then $P_2(\Omega) =P_2'(\Omega)+ P_2''(\Omega) $ shows that $P_2''(\Omega) =0$. Thus $P_2''$ is the zero measure. So $P_2=P_2' \perp P_1$.