Do limits respect measure?

Question

I have that in a metric space $X$, with sets forming a $\sigma$-algebra $$A_1, A_2, \cdots \to A \ne \{\emptyset\} $$ such that $$A_1 \supset A_2 \supset \cdots$$ Here's what I want to say, but I'm not sure if I can or not: $$\lim_{i \to \infty}A_i=A \Rightarrow \mu(\lim_{i \to \infty}A_i)=\mu(A) \Rightarrow \lim_{i \to \infty}\mu(A_i)=\mu(A)$$ Is this an acceptable statement? Or do limits not necessarily respect measure?

Answer 1

Let $B_n = A_n-A_{n+1}$. Then we have (1) $B_n$ are disjoint (2) $\bigcup_{n=1}^{\infty}B_n = A_1-A$. Thus $$\mu(A_1)-\mu(A)=\sum_{n=1}^{\infty}\mu(B_n)=\sum_{n=1}^{\infty}\left(\mu(A_n)-\mu(A_{n+1})\right) = \mu(A_1)-\lim_{N\to\infty}\mu(A_N)$$ so that if $\mu(A_1)$ is finite, we can subtract it to have $\mu(A) = \lim_{N\to\infty}\mu(A_N)$. If $\mu(A_1)$ is allowed to be $\infty$, then the answer is negative. One may consider $\mathbb{R}$ with Lebesgue measure and let $A_n = (n, \infty)$. $\lim\mu(A_n) = \infty$ but $\mu(A) = 0$.

Answer 2

Not in this generality.

For instance consider the measure $\mu$ on $[0,1]$ which assigns a set $A$ the measure $\sum_{n=1}^\infty \delta_n(A)$ where $$\delta_n(A) = \begin{cases}1 & \frac{1}{n}\in A \\ 0 & \text{otherwise}\end{cases}$$

Now look at $A_m = (0,\frac{1}{m})$. Then since $A_m$ contains infinitely many elements in the set $\{\frac{1}{n} : n\in \mathbb{N}\}$, the measure must be infinity. However $\lim_{m\rightarrow\infty} A_m =\emptyset$ has measure zero.

Let me note that there are many instances where this property does hold (I would google regular measures as a start)

Answer 3

The answer is yes, if at least one of the sets $A_{i}$ has finite measure (or if the measure is finite altogether, e.g. a probability measure). This property is known as "continuity from above", see here (scroll down to section "Continuity from above", where you also find a counterexample if the finiteness assumption is violated).