Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need help with this problem:

Let $\mathcal{M}$ be a sigma algebra of subsets of $X$ and the set function $\mu:\mathcal{M}\to [0,\infty)$ be finitely additive. Prove that $\mu$ is a measure if and only when $\{A_k\}_{k=1}^\infty$ is a descending sequence of sets in $\mathcal{M}$, then $$ \mu\left(\bigcap_{k=1}^\infty A_k\right) = \lim_{k\to \infty} \mu(A_k).$$

Below is my attempt:

Clearly one direction follows from continuity of measure.

Now suppose that the display equation above is true. Let $E=\cup_{k=1}^\infty E_k$, where the $E_k$ are disjoint. Set $A_k=E\setminus \cup_{k=1}^n E_k$. Then $A_k \supseteq A_{k+1}$. Also, $\cap_{k=1}^\infty A_k = \emptyset$. $\mu$ is also finitely additive, so $$ \mu(A_k)= \mu(E)- \sum_{n=1}^k \mu(E_n).$$ So $$\begin{align*} \mu(E) & = \lim_{k\to \infty} \mu(A_k)+\lim_{k\to \infty} \sum_{n=1}^k \mu(E_n)\\ & = \mu \left(\cap_{k=1}^\infty A_k\right) + \sum_{k=1}^\infty \mu(E_k)\\ & = \sum_{k=1}^\infty \mu(E_k) \end{align*} $$

Please, does this look right?

share|cite|improve this question
There is typo in the line before the last, it should be $\mu(\bigcap_{k=1}^{+\infty}A_k)$. I think you have justify that $\mu(\emptyset)=0$ (even if it's obvious). Note that the fact that the measure is finite is needed to get the first direction, so it may be more detailed. – Davide Giraudo Feb 20 '12 at 14:19
Thanks. I've fixed it. So If I justify that $\mu(\emptyset)=0$, it'll be fine? – Bill Feb 20 '12 at 14:24
And if you give more details for the first direction, yes! – Davide Giraudo Feb 20 '12 at 14:26
ok. Thanks very much. – Bill Feb 20 '12 at 14:28
You're welcome. – Davide Giraudo Feb 20 '12 at 14:28
up vote 2 down vote accepted

First, we assume that $\mu$ is a measure. Let $\{A_k\}\subset \mathcal M$ a decreasing sequence of measurable subsets of $X$. Put $B_k:=X\setminus A_k$ for $k\geq 0$ and $C_k:=B_k-B_{k-1}$ for $k\geq 1$, $C_0:=B_0$. Then the sets $C_k$ are pairwise disjoint, $\bigcup_{k\geq 0}C_k=\bigcup_{k\geq 0}B_k=\bigcup_{k\geq 0}A_k^c=X\setminus \bigcap_{k\geq 0}A_k$ and since $\mu(B_k)$ is finite we have $\mu(C_k)=\mu(B_k)-\mu(B_{k-1})$. Hence $$\mu\left(\bigcap_{k\geq 0}A_k\right)=\mu(X)-\mu\left(X\setminus \bigcap_{k\geq 0}A_k\right)=\mu(X)-\mu\left(\bigcup_{k\geq 0}C_k\right)\\\ =\mu(X)-\sum_{k\geq 1}(\mu(B_k)-\mu(B_{k-1}))-\mu(B_0)=\mu(X)-\lim_{k\to \infty}\mu(B_k)=\lim_{k\to \infty}\mu(A_k).$$

For the converse what you did is fine,I would just add that $\mu(\emptyset)=\mu(\emptyset\cup\emptyset)=\mu(\emptyset)+\mu(\emptyset)$ and since $ \mu(\emptyset)<\infty$ we have $\mu(\emptyset)=0$.

The fact that the measure was finite was necessary here, for example if we take $\mu(A)=+\infty$ for all $A\in\mathcal M$ we won't get a measure since we need $\mu(\emptyset)=0$.

share|cite|improve this answer
Every measure satisfies $\mu(\emptyset)=0$. But the finiteness of measures is necessary. Lebesgue measure and the family of sets $[n.\infty)$ provide a counterexample that the condition doesn't work for infinite measures. – Michael Greinecker Feb 20 '12 at 14:49

The assumptions is false $A_{k}=[k,\infty)$ is decreasing however the measure of the intersection is zero when the measure of every set is $\infty$. You have to assume that the measure of one set is not $\infty$.

share|cite|improve this answer
It's the case since the range of $\mu$ is supposed to be contained on $[0,+\infty)$. – Davide Giraudo Feb 20 '12 at 18:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.