# Prove the measure of limit equals the limit of measure

If ($$\Omega, \mathscr{A}, \mu$$) is a measure space, and $$\forall A_n \in \mathscr{A}$$

Prove that

$$\mu(\liminf_{n\rightarrow\infty}A_n) = \lim_{n\rightarrow\infty}\mu(\cap^\infty_{i=n}A_i)$$

Moreover, Suppoes that $$\mu(\cap^\infty_{i=n}A_i)<\infty$$ for some $$n \geq 0$$. Prove that

$$\mu(\limsup_{n\rightarrow\infty}A_n) = \mu(\cup^\infty_{i=n}A_i)$$

My major confusion is the monotonicity of $$A_n$$ is not given.

I tried this approach(monotone continuity from below) and substitute its $$A_n$$ with $$\cap^\infty_{i=n}A_i$$ but I don't think it works.

Please inspire me, I've been working on this for the whole day.

Your idea is correct. Use continuity of the measure from below applied to the sequence $$B_n = \bigcap_{j = n}^{\infty}A_j$$. The full solution is in the next paragraph:
By definition, $$\liminf_{n \to \infty}A_n = \bigcup_{n = 1}^{\infty}\bigcap_{j = n}^{\infty}A_j$$. Hence $$\bigcap_{j = n}^{\infty}A_j \nearrow \liminf_{n \to \infty} A_n$$ as $$n \to \infty$$. By continuity of measure from below, $$\mu(\bigcap_{j = n}^{\infty}A_j) \nearrow \mu(\liminf_{n \to \infty} A_n)$$ as $$n \to \infty$$.