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Background information - Let $X$ be a set well equipped with a $\sigma$-algebra $M$. A measure on $M$ (or on $(X,M)$ or on $X$ if $M$ is understood) is a function $\mu: M \rightarrow [0,\infty]$ such that

i.) $\mu(\emptyset) = 0$

ii.) if $\{E_j\}_{1}^{\infty}$ is a sequence of disjoint sets in $M$ then $\mu(\cup_{1}^{\infty}E_j) = \sum_{1}^{\infty}\mu(E_j)$

Property (ii.) is called countable additivity and implies finite additivity:

ii.*) if $E_1,\ldots, E_n$ are disjoint sets in $M$ then $\mu(\cup_{1}^{n}E_j) = \sum_{1}^{n}\mu(E_j)$

because one can take $E_j = \emptyset$ for $j >n$. A function $\mu$ that satisfies (i.) and (ii.*) but not (ii.) is called a finitely additive measure.

A finitely additive measure $\mu$ is a measure if and only if it is continuous from below. If $\mu(X) < \infty$, $\mu$ is a measure if and only if it is continuous from above.

The proof of the first part is provided in the answer here

Attempted proof of second part - Let $\{E_j\}_{1}^{\infty}$ be a sequence of disjoint sets in $M$ and $\mu(X) < \infty$. Let $$F_k = \bigcap_{1}^{\infty}E_j$$ As you can see $F_1\supset F_2 \supset \ldots$, so the $F_k$'s are a decreasing sequence of sets and clearly $$\bigcap_{1}^{\infty}F_k = \bigcap_{1}^{\infty}E_j$$ Now from continuity from above $$\mu\left(\bigcap_{1}^{\infty}E_j\right) = \mu\left(\bigcap_{1}^{\infty}F_k\right) = \lim_{k\rightarrow \infty}\mu(F_k)$$ Now what I want to do is apply De'Morgans law and conclude that we will have continuity from above being the same as continuity from below but I am not sure if that I can actually do that. Any suggestions is greatly appreciated.

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  • $\begingroup$ If the $E_j$'s are disjoint then each $F_k$ is empty. $\endgroup$ – Umberto P. May 16 '16 at 14:05
  • $\begingroup$ Ok so I should not make the $E_j$'s disjoint then? $\endgroup$ – Wolfy May 16 '16 at 14:07
  • $\begingroup$ That depends on what you are trying to prove. $\endgroup$ – Umberto P. May 16 '16 at 14:07
  • $\begingroup$ Is it not clear what I am trying to prove? $\endgroup$ – Wolfy May 16 '16 at 14:08
  • $\begingroup$ No, not really. Perhaps you should write out the condition you need to verify for $\mu$ to be a measure. $\endgroup$ – Umberto P. May 16 '16 at 14:09
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Let $\mu$ be a finitely additive measure such that $\mu(X) < \infty$. Then $\mu$ is a measure if and only if it is continuous from above.

Proof:

($\Rightarrow$) Since $\mu$ is a measure, we know it is continuous from below. Now, let $\{F_j\}_{1}^{\infty}$ be a non-increasing sequence of sets in $M$. Define $E_k=X \setminus F_k$. Then $\{E_k\}_{1}^{\infty}$ is a non-decreasing sequence of sets in $M$, and $\cap_{j=1}^{\infty}F_j=X \setminus \cup_{k=1}^{\infty}E_k$, so we have, using the fact that $\mu(X) < \infty$, \begin{align} \mu(\cap_{j=1}^{\infty}F_j)&= \mu(X \setminus \cup_{k=1}^{\infty}E_k)=\mu(X)-\mu(\cup_{k=1}^{\infty}E_k) =\mu(X) -\lim_{k\to \infty}\mu(E_k)=\\ &=\mu(X) -\lim_{k\to \infty}\mu(X \setminus F_k)=\mu(X) -\lim_{k\to \infty}(\mu(X)-\mu(F_k))= \lim_{k\to \infty}\mu(F_k) \end{align} So $\mu$ is continuous from above.

($\Leftarrow$) Let $\{E_j\}_{1}^{\infty}$ is a sequence of disjoint sets in $M$. Define $F_k= \bigcap_{j=1}^k (X \setminus E_j)$. Then $\{F_k\}_{1}^{\infty}$ is a non-increasing sequence of sets in $M$ and $\cap_{k=1}^{\infty}F_k=X \setminus \cup_{j=1}^{\infty}E_j$, so we have, using the fact that $\mu(X) < \infty$,

\begin{align} \mu(X) - \mu(\cup_{j=1}^{\infty}E_j)&=\mu(X \setminus \cup_{j=1}^{\infty}E_j) = \mu(\cap_{k=1}^{\infty}F_k)= \lim_{k \to \infty} \mu(F_k)= \\ &=\lim_{k \to \infty} \mu( \cap_{j=1}^{k}(X \setminus E_j))=\lim_{k \to \infty} \mu(X \setminus \cup_{j=1}^{k}E_j)= \\ &= \lim_{k \to \infty} (\mu(X)-\mu(\cup_{j=1}^{k}E_j))=\mu(X)-\lim_{k \to \infty} \mu(\cup_{j=1}^{k}E_j)= \\ & =\mu(X)-\lim_{k \to \infty} \sum_{j=1}^{k} \mu(E_j)= \mu(X)-\sum_{j=1}^{\infty} \mu(E_j) \end{align} Using againg that $\mu(X) < \infty$, we have $$\mu(\cup_{1}^{\infty}E_j) = \sum_{1}^{\infty}\mu(E_j)$$

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Hints:

1). $X\setminus \left( \bigcup_{j=1}^\infty E_j \right)=\bigcap_{j=1}^\infty(X\setminus E_j) $

2). $\mu (X\setminus E_j)=\mu (X)-\mu (E_j);\ (\mu$ is finite).

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Work with $F_i=E_i^C$ then $\mu(\cap F_i)=\mu(X)-\lim\mu(E_i)$ by cont from above but the left hand side is $\mu( (\cup E_i)^C)$ by de Morgan and so is $\mu(X)-\mu(\cup E_i)$ and you are done.

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