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A finitely additive measure $\mu$ is a measure if and only if it is continuous from below.

I want to know how I should proceed in proving this statement. My idea is to first assume we have a finitely additive measure $\mu$ and then prove that it is continuous from below. Then for the converse assume we have an increasing sequence of sets in $M$ and show that $\mu$ is finitely additive.

I am not sure if this is the correct approach I should take. Any suggestions is greatly appreciated.

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    $\begingroup$ You don't need to prove that $\mu$ is finitely additive. The problem statement means: Given a finitely additive measure $\mu$, prove: $$ \mu \textrm{ is a measure } \Leftrightarrow \; \mu \textrm{ is continuous from below }$$ $\endgroup$ – Ramiro May 14 '16 at 18:04
  • $\begingroup$ I see understand it now thank you $\endgroup$ – Wolfy May 14 '16 at 22:47
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Note that the only difference between a measure and a finitely additive measure is that a finitely additive measure may not satisfy countable additivity. Moreover, if $\mu$ is in fact a measure then it is a standard fact that $\mu$ satisfies continuity from below. So what the problem is really asking is to show that a finitely additive measure which satisfies continuity from below is countably additive, hence is a measure.

To show this, let $\{E_n\}_{n=1}^{\infty}$ be a sequence of disjoint sets in the $\sigma$-algebra $\mathcal{M}$. Define $$ F_k=\bigcup_{n=1}^kE_n$$ and observe that $F_1\subset F_2\subset\cdots$ and that $$ \bigcup_{n=1}^{\infty}E_n=\bigcup_{k=1}^{\infty}F_k$$ Therefore by finite additivity and continuity from below, $$\mu\Big(\bigcup_{n=1}^{\infty}E_n\Big)=\mu\Big(\bigcup_{k=1}^{\infty}F_k\Big)=\lim_{k\to\infty}\mu(F_k)=\lim_{k\to\infty}\sum_{n=1}^k\mu(E_n)=\sum_{n=1}^{\infty}\mu(E_n)$$

Therefore $\mu$ is countably additive, hence is a measure.

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  • $\begingroup$ Can you just show me how $F_1\subset F_2\subset\ldots$? $\endgroup$ – Wolfy May 15 '16 at 12:31
  • $\begingroup$ Also, do we apply the same idea for continuity from above? $\endgroup$ – Wolfy May 15 '16 at 12:32
  • $\begingroup$ $F_k=E_1\cup\dots\cup E_k$ and $F_{k+1}=E_1\cup\cdots \cup E_k\cup E_{k+1}=F_k\cup E_{k+1}$, so $F_k\subset F_{k+1}$. A similar idea should work for continuity from above, perhaps with some complements. $\endgroup$ – carmichael561 May 15 '16 at 15:53
  • $\begingroup$ Should I re-edit my answer to include continuity from above or just make a new question? $\endgroup$ – Wolfy May 16 '16 at 13:37
  • $\begingroup$ I think a new question would be better. $\endgroup$ – carmichael561 May 16 '16 at 15:18

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