For a set $X$ with a $\sigma$-algebra $\xi \subseteq \mathcal{P}(X)$ and $\sigma$-additive $\mu: \xi \rightarrow [0, \infty]$. The following inequality holds for $(A_n)_{n \in \mathbb{N}} \in \xi^\mathbb{N}$:
$$ \mu({\liminf}_{n \rightarrow \infty} A_n) \leq {\liminf}_{n \rightarrow \infty} \mu(A_n) $$
and under the assumption of $\mu(X) < \infty$ we also get:
$$ {\limsup}_{n \rightarrow \infty} \mu(A_n)\leq \mu({\limsup}_{n \rightarrow \infty} A_n) $$
I was able to prove these inequalities, but can't seem to develop a deeper intuition. Is there any way to better understand why these statements are true?
Well, let's focus on the first inequality. The idea is simple: on the left-hand-side one is taking the intersection of the sets from some moment, while on the right-hand-side one is taking the smallest size of the sets from some moment. If you consider $X := X_1\cap \cdots \cap X_n$, there is no way that $|X| > |X_i|$ for some $i$, right? The same reason here, and all you need to do is to fit it in a limit and measure theoretic setting.