Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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 am trying to prove every $\sigma$-finite measure is semifinite. This is what I have tried:

Definition of $\sigma$-finiteness: Let $(X,\mathcal{M},\mu)$ is a measure space. Then, for $E_i \in \mathcal{M}$, $X = \bigcup_{i=1}^{\infty}E_i$ where $\mu(E_i) < \infty$.

Definition of semifiniteness: For each $E \in \mathcal{M}$ with $\mu(E) = \infty$ $\exists$ $F \subset E$ and $F \in \mathcal{M}$ and $0 < \mu(F) < \infty$.

So, take $A$ s.t. $\mu(A) = \infty$. We know $X \cap A = A$. Then, $A = A \cap \bigcup E_j$ hence $A = \bigcup E_j \cap A$. By subadditivity,

$$\infty = \mu(A) = \mu\left(\bigcup E_j \cap A\right) \leq \sum_1^{\infty} \mu(E_j \cap A) $$

OK, I am here. But I do not understand how to continue, or even this is a right approach. Thanks.

share|cite|improve this question
I think you meant $F\subseteq E$ in the definition of semifinite. Also there are something off in the definition of $\sigma$-finite (e.g. "There are $E_i$ such that...). – Asaf Karagila Nov 11 '12 at 14:30
Yes, Asaf, thanks. – oeda Nov 11 '12 at 14:31
The $E_i$ are not included in $\cal M$, but are elements of $\cal M$. – Davide Giraudo Nov 11 '12 at 15:11
Yes, okay, this was another typo, corrected it, thanks. – oeda Nov 11 '12 at 15:37
up vote 4 down vote accepted

We can find $n$ such that $\mu\left(A\cap\bigcup_{j=1}^NE_j\right)>0$, and we have $\mu\left(A\cap\bigcup_{j=1}^NE_j\right) < \mu(A) < \infty$. Furthermore, $A\cap\bigcup_{j=1}^NE_j\subset A$, so $\mu$ is semi-finite.

The converse is not true: counting measure on the subsets of $[0,1]$ is semi-finite but not $\sigma$-finite.

share|cite|improve this answer

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.