Last year I gave a proof of a standard inequality regarding the limit superior and limit inferior of a sequence of sets that was, quite frankly, not elegant and incomplete at best. I am going to fix this here. I will also generalize the result to finite measures.

Proposition: Let $\{A_n\}_{n=1}^{\infty}$ be a sequence of measurable sets, and let $\mu$ be a finite measure. We have the following inequalities, $$ \begin{align*} \mu(\liminf_{n\rightarrow \infty} A_n) \leq \liminf_{n \rightarrow \infty} \mu(A_n) \leq \limsup_{n \rightarrow \infty} \mu(A_n) \leq \mu(\limsup_{n \rightarrow \infty} A_n) \:. \end{align*} $$

Proof: Recall that $$ \begin{align*} \liminf_{n\rightarrow \infty} A_n := \bigcup_{n \geq 1} \bigcap_{k \geq n} A_k, \qquad \limsup_{n\rightarrow \infty} A_n := \bigcap_{n \geq 1} \bigcup_{k \geq n} A_k \:. \end{align*} $$ Let us work on the LHS inequality first. Define $B_n := \bigcap_{k \geq n} A_k$ for $n \geq 1$. Observe that $B_1 \subset B_2 \subset ...$. Hence, by continuity of measure from below, we have $$ \begin{align*} \mu(\bigcup_{n \geq 1} \bigcap_{k \geq n} A_k) = \mu(\bigcup_{n \geq 1} B_n) = \lim_{n \rightarrow \infty} \mu(B_n) = \lim_{n \rightarrow \infty} \mu( \bigcap_{k \geq n} A_k ) \:. \end{align*} $$ Now, by monotonicity of measure, for any $k \geq n$, we have $\mu(\bigcap_{k \geq n} A_k) \leq \mu(A_k)$. Since $k \geq n$ is arbitrary, we can take the infimum on the RHS to conclude $\mu(\bigcap_{k \geq n} A_k) \leq \inf_{k \geq n} \mu(A_k)$. But this means that $$ \begin{align*} \lim_{n \rightarrow \infty} \mu(\bigcap_{k \geq n} A_k) \leq \liminf_{n \rightarrow \infty} \mu(A_n) \:, \end{align*} $$ which proves the LHS inequality.

The RHS inequality proceeds similarly. Define $C_n := \bigcup_{k \geq n} A_k$ for $n \geq 1$. Now observe that $C_1 \supset C_2 \supset ...$. Since we assumed that $\mu$ is finite, we have $\mu(C_1) < \infty$. Hence, by continuity of measure from above, we have $$ \begin{align*} \mu(\bigcap_{n \geq 1} \bigcup_{k \geq n} A_k) = \mu(\bigcap_{n \geq 1} C_n) = \lim_{n \rightarrow \infty} \mu(C_n) = \lim_{n \rightarrow \infty} \mu( \bigcup_{k \geq n} A_k ) \:. \end{align*} $$ Once again by monotonicity of measure, $\mu(\bigcup_{k \geq n} A_k) \geq \mu(A_k)$ for any $k \geq n$, which means that $\mu(\bigcup_{k \geq n} A_k) \geq \sup_{k \geq n} \mu(A_k)$. This shows that $$ \begin{align*} \lim_{n \rightarrow \infty} \mu(\bigcup_{k \geq n} A_k) \geq \limsup_{n \rightarrow \infty} \mu(A_n) \:, \end{align*} $$ which proves the RHS inequality.