TITLE: Measure theory: existence of a monotone sequence of sets
QUESTION [7 upvotes]: Let $(X,\mathfrak B,\mu)$ be a probability space, and suppose that $\lim_n \mu(A_n) = m$ for some sequence of measurable sets $\{A_n\}_{n\geq 0}\subseteq \mathfrak B$. Is it true that there exists a monotone sequence $\{B_n\}_{n\geq 0}\subseteq \mathfrak B$ that satisfies  $\lim_n \mu(B_n) = m$?
The sequence $\{B_n\}_{n\geq 0}$ is monotone if either $B_n\subseteq B_{n+1}\; \forall n$, or $B_n\supseteq B_{n+1}\; \forall n$.

Motivation is taken from this question: existence of $\{B_n\}_{n\geq 0}$ implies that $\mu(\lim_n B_n) = m$, and thus the image of $\mu$ is closed in $\Bbb R$. 
Using the fact that any sequence of reals contains a monotone subsequence (non-increasing, or non-decreasing, or both), we can assume that $\{\mu(A_n)\}_{n\geq 0}$ is a monotone sequence. For such case, I tried to choose $B_n$ using $\limsup/\liminf$-like constructions, but did not obtain any particular result.

REPLY [2 votes]: As it follows from the answer of Michael, the range of $\mu$ is closed and thus there exists a set $B$ such that $\mu(B) = m$. If one takes $B_n\equiv B$, the sequence is obviously monotone, so the answer to my question is indeed positive.