While studying measures I was wondering about the following:
Let $(X,\mathcal{A},\mu)$ be a measure space and assume we have a sequence of sets $\{A_n\}_{n\in\mathbb{N}} \subset \mathcal{A}$ with $A_{n+1} \subseteq A_{n}$ such that $\mu(A_n) = 1$ for all $n\in \mathbb{N}$, can we conclude that the following statement is always true:
$\mu(\bigcap_{n\in\mathbb{N}} A_n) =\lim_{n \to \infty}\mu(A_n) = 1 $ ?
I would say that it is, since the function $f(n):= \mu(A_n) = 1$ is constant over $\mathbb{N}$ but I'm not sure. Some help would be highly appreciated
I'm re-editing this question because I think I actually came up with a proof:
We first define $I_n:= A_0\setminus A_n$. Then we have that:
\begin{equation} \begin{split} \mu(\bigcup_{n\in \mathbb{N}} I_n) = \mu(\bigcup_{n\in \mathbb{N}}( A_0\setminus A_n ) &\quad \\= \mu(A_0\setminus\bigcap_{n\in \mathbb{N}}A_n) = \mu(A_0) - \mu(\bigcap_{n\in \mathbb{N}}A_n) = \mu(A_0) - \lim_{n\to \infty}\mu(A_n) \\ \leq \sum_{n\in\mathbb{N}}\mu(\ A_0\setminus A_n ) &\quad \\= \sum_{n\in\mathbb{N}} \mu(\ A_0) - ( A_n ) &\quad \\= \sum_{n\in\mathbb{N}} (1-1) &\quad \\= 0 . \end{split} \end{equation} Finally $\mu(A_0) - \lim_{n\to \infty}\mu(A_n) = 0$ implies that $\lim_{n\to \infty}\mu(A_n) =\mu(A_0) = 1$
Does that make sense?
Under your monotonicity assumption ($A_{n+1} \subset A_n$ for every $n$) and the assumption that $\mu(A_N) < +\infty$ for some $N$ (and hence for every $n\geq N$), you always have that $$ \mu\left(\bigcap_n A_n\right) = \lim_n \mu(A_n). $$
(See for example here for the analogous result for increasing sequences of sets.)