I have the following theorem:

Theorem: Let$(S,\mathcal{A},\mu)$ be a measurable space and let $(A_n)_{n\geq 1}$ be a sequence in $\mathcal{A}$.

i) If $A_n \uparrow A$, then $\mu(A_n) \uparrow \mu(A)$.

ii) If $A_n \downarrow A$ and $\mu(A_1)<\infty$, then $\mu(A_n)\downarrow \mu(A)$.

(Note that in this definition $A = \bigcup\limits_{n=1}^{\infty}A_n$).

Exercise: Prove i) and ii).

What I've tried:

i) Pick a disjoint sequence $(B_n)_{n\geq 1}$. We have that $\bigcup\limits_{j=1}^{\infty}B_n = A$ and $\bigcup\limits_{j=1}^{n}B_n = A_j$. Therefore, $\sigma$-additivity gives$$\mu(A) = \sum\limits_{j =1}^{\infty}\mu(B_j) = \lim\limits_{n\to\infty}\sum\limits_{j = 1}^n \mu(B_j) = \lim\limits_{n\to\infty}\mu(A_n).$$ We have that $\mu(A) = \lim\limits_{n\to\infty}\mu(A_n)$ implies that $\mu(A)\geq \mu(A_n)$.

ii) I'm not sure what to do here. Again, I pick a disjoint sequence $(B_n)_{n\geq1}$. We have that $\bigcap\limits_{j=1}^\infty B_j = A$ and $\bigcap\limits_{j=1}^n B_j = A_n$. This time I need to show that $\mu(A_n)\geq \mu(A)$, right? I think this is equivalent to showing that $\bigcap\limits_{j=1}^\infty \mu(A_j) = \mu(A)$. However, I don't know how to use $\sigma$-additivity in this case.

Question: How do I show that if $A_n \downarrow A$ and $\mu(A_n)<\infty$, then $\mu(A_n)\downarrow \mu(A)$?

Thanks in advance!

  • $\begingroup$ Do you mean $\mu(A_1) \lt \infty$ in the question? $\endgroup$ – tattwamasi amrutam Nov 26 '17 at 20:19
  • $\begingroup$ @tattwamasiamrutam Yes, thank you! $\endgroup$ – titusAdam Nov 30 '17 at 13:20
  • $\begingroup$ Take $B_n=A_1\A_n$ and use part (I) $\endgroup$ – tattwamasi amrutam Nov 30 '17 at 13:22
  • $\begingroup$ for (ii) you can use (i) by taking complements. Also in (ii) $A$ should be $\cap A_j$ $\endgroup$ – daw Nov 30 '17 at 17:57

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