## Probability Foundations – Problem (39/365)

Let $\mu$ be a finite measure on algebra $\mathcal{A}$, $A_n \in \mathcal{A}$ for $n = 1,2,\dots$ and $A = \lim_n A_n$ (i.e. $A=\overline{\lim}A_{n}=\underline{\lim}A_{n}$). Show that $\mu(A) = \lim_n \mu(A_n$.

$\displaystyle \overline{\lim}A_{n} = \cap_{n=1}^{\infty}\cup_{k=n}^{\infty}A_{k} \\ = \left(\cup_{k=1}^{\infty}A_{k}\right)\cap\left(\cup_{k=2}^{\infty}A_{k}\right)\cap\dots \\ = \left(\cup_{k=2}^{\infty}A_{k}\right)\cap\dots \\ = \lim_{n}\left(\cup_{k=n}^{\infty}A_{k}\right) \\ \underline{\lim}A_{n} = \cup_{n=1}^{\infty}\cap_{k=n}^{\infty}A_{k} \\ = \left(\cap_{k=1}^{\infty}A_{k}\right)\cup\left(\cap_{k=2}^{\infty}A_{k}\right)\cup\dots \\ = \left(\cap_{k=2}^{\infty}A_{k}\right)\cup\dots \\ = \lim_{n}\left(\cap_{k=n}^{\infty}A_{k}\right)$

Since $A=\overline{\lim}A_{n}=\underline{\lim}A_{n}$,

$\displaystyle \lim_{n}\left(\cup_{k=n}^{\infty}A_{k}\right) = \lim_{n}\left(\cap_{k=n}^{\infty}A_{k}\right) \\ \text{implies limit is } \lim_n A_n \text{ because the union and intersection have to agree}\\ \text{so } \mu(A) = \lim_n \mu(A_n)$

This entry was posted in Uncategorized and tagged , . Bookmark the permalink.