Let $P$ be a Borel probability measure over $(X, \mathcal{X})$, where $X\subset \mathbb{R}^d$, for some $d \geq 1$. Assume $P$ is absolutely continuous with respect to the Lebesgue measure $\lambda$ on $(X, \mathcal{X})$. Let $(B_n) \subset \mathcal{X}$, is the implication $$ \lim_{n\to \infty} \lambda(B_n)=0 \quad \implies \lim_{n\to \infty}P(B_n)=0 $$ correct?
I would say: yes. By definition of absolute continuity, $\forall\epsilon>0$ there exists $\delta$ such that $P(B)<\epsilon$ for all Borel sets $B$ such that $\lambda(B)<\delta$. In particular, there exists $n_\delta$ such that $\lambda_n(B_n)<\delta$ for all $n \geq n_\delta$ ($B_n$ is as above). Now, if $P(B_n)$ did not converge to $0$, there would exist $n \geq n_\delta$ such that $P(B_n)>\epsilon$, yelding a contradiction. Is my reasoning correct? I guess it carries over to the more general case of absolutely continuous Borel measures.
