Take the 2-minute tour ×
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.

Let $(X,M,\mu)$ be a measure space. Why are the following statements equivalent?

i) There exists a sequence of pairwise disjoint measurable sets $\{A_{n}\}$ in $M$ , each of finite measure, such that $\mu(A) = \sum_{n=1}^{\infty} \mu(A \cap A_{n})$ for every measurable set $A \in M$.

ii) $\mu$ is a countable sum of pairwise mutually singular finite measures.

share|improve this question
For the future, you might consider editing your title to be more informative. That way, you're more likely to get more views and more answers. –  Jesse Madnick Dec 6 '10 at 10:31

1 Answer 1

(ii) $\implies$ (i) Suppose $\mu = \sum_{n=1}^\infty \mu_n$, where $\mu_i \perp \mu_j$ for all $i \neq j$. By definition of "mutually singular," there exist pairwise disjoint $A_n \in M$ such that $\mu_n(A) = \mu_n(A \cap A_n)$ for all $A \in M$. Therefore, $\mu(A_k) = \sum_{n=1}^\infty \mu_n(A_k) = \mu_k(A_k) < \infty$, so that each $A_n$ has finite $\mu$-measure. Moreover, $\mu(A) = \sum_{n=1}^\infty \mu_n(A_n) = \sum_{n=1}^\infty \mu_n(A \cap A_n) = \sum_{n=1}^\infty \mu(A \cap A_n)$, where in the last step we used the fact that $\mu(A \cap A_n) = \mu_n(A \cap A_n)$ for all $n \in N$.

(i) $\implies$ (ii) Let $\mu_n$ be a measure concentrated on $A_n \in M$, normalized so that $\mu_n(A_n) = 1$. Then $\mu(A) = \sum_{n=1}^\infty \mu(A\cap A_n) = \sum_{n=1}^\infty \mu_n(A)$.

(EDIT following Nate's comment: We can define $\mu_n(A) = \mu(A \cap A_n)$.)

share|improve this answer
For (i) $\implies$ (ii), can't we just let $\mu_n(A) = \mu(A \cap A_n)$? –  Nate Eldredge Dec 6 '10 at 14:08
Oh, hah, right: it is, in fact, possible to take the measure of empty sets. Yes, that does simplify things... –  Jesse Madnick Dec 6 '10 at 18:39

Your Answer


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