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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.

Measure Theory
and Probability Theory by Krishna B. Athreya
Soumendra N. Lahiri page 15

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1 Answer 1

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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 $n \to \infty$.

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  • $\begingroup$ Thank you very much $\endgroup$ Commented Oct 18, 2021 at 1:01

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