If $(\Omega, \mathcal{A}, \mu)$ is a measure space, and $A_{n} \in \mathcal{A},$ then $\mu(\varliminf _{n \rightarrow \infty} A_{n}) \leq \varliminf_{n \rightarrow \infty} \mu(A_{n})$
I tried to prove the statement by going back to the definition of liminf of a set:
$$\mu(\varliminf A_n)=\mu(\bigcup^\infty_{n=1}\bigcap^\infty_{n=k}A_n)$$
As I know $\bigcap^\infty_{n=k}A_n$ is increasing when $k$ increases, I get $$\mu(\bigcup^\infty_{n=1}\bigcap^\infty_{n=k}A_n)=\lim\mu(\bigcap^\infty_{n=k}A_n)$$ but I am stucked from here. Intuitively, $\bigcap^N_{n=k}A_n$ is decreasing when $N\rightarrow\infty$ but as $n$ starts from $k$, therefore it seems $\mu(\bigcap^N_{n=k}A_n)$ is bounded above but how do I connect it to the $\inf \mu(A_n)$? On the other hand, it seems $\bigcap^\infty_{n=k}A_n$ is a decreasing sequence that is bounded below by $n\geq k$. I am so confused with the bounded above and bounded below and the infimum in this case. I hope my question make sense. Could someone please explain to me how should I look at this question and reason the statement rigorously?
Best Answer
As you have written, $$\mu(\liminf_{n\to\infty} A_n) = \lim_{k\to\infty} \mu\left(\bigcap_{n=k}^\infty A_n\right)$$ where the limit on the right exists (i.e. it is in particular equal to $\liminf_{k\to\infty} \mu\left(\bigcap_{n=k}^\infty A_n\right)$). As @Jakobian was saying, $a_k:=\mu\left(\bigcap_{n=k}^\infty A_n\right) \leq \mu(A_k) =: b_k$. Now if we have two sequences of real numbers $\{a_k\}_{k=1}^\infty$ and $\{b_k\}_{k=1}^\infty$ where each $a_k \leq b_k$, then the $\liminf$ of the $a_k$'s MUST be $\leq $ the $\liminf$ of the $b_k$'s, and so we can finish the above chain of equalities: $$\mu(\liminf_{n\to\infty} A_n) = \lim_{k\to\infty} \mu\left(\bigcap_{n=k}^\infty A_n\right) = \liminf_{k\to\infty} \mu\left(\bigcap_{n=k}^\infty A_n\right) = \liminf_{k\to\infty}a_k \leq \liminf_{k\to\infty}b_k = \liminf_{k\to\infty}\mu(A_k)$$