Probability Foundations – Problem (39/365)

Posted on October 20, 2016 by Naren Sundar

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)

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