# $\mu$ on $\mathcal{A}$ is $\sigma$ finite if and only if $\mu$ on $R$ is $\sigma$ finite

I have been struggling with the following problem for many hours now :

Suppose $R$ is an algebra of sets on $X$ and $\mathcal{A}$ is the $\sigma$-algebra generated by $R$. Let $\mu$ be a measure defined on $\mathcal{A}$. Show that $\mu$ is $\sigma$-finite on $\mathcal{A}$ if an only if it is $\sigma$-finite on $R$.

One of the implications is trivial, since if you have a collection of sets in $R$ they are also in $\mathcal{A}$.

The other one, I have been struggling with.

If $\mu$ was actually constructed from $\mu$ on $R$ using the outer measure construction, than the solution would be easy. The problem is, we know nothing about $\mu$ on $\mathcal{A}$ a priori.

In order to show that $\mu$ on $\mathcal{A}$ is equal to $\mu^*$, the outer measure constructed via $\mu$ on $R$, you would need to use the fact that $\mu$ is $\sigma$-finite on $R$, which is precisely what we are trying to prove...

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I think the following lemma can be useful: If $(X,\mathcal B,\mu)$ is a finite measure space, and $\mathcal A$ is an algebra with generates $\mathcal B$ then for each $B\in\mathcal B$ and $\varepsilon>0$, we can find $A\in\mathcal A$ such that $\mu(A\Delta B)\leq \varepsilon$. – Davide Giraudo Sep 22 '12 at 13:39
Is it an exercise? Where is the problem from? – Davide Giraudo Sep 23 '12 at 14:08
It is a problem in Benedetto and Czaja "Integration and Modern Analysis". I think I found a solution to the problem now, should I post it as an answer here myself? – Vhailor Sep 23 '12 at 15:49
Yes, you should. It's allowed and welcomed. – Davide Giraudo Sep 23 '12 at 17:08
Did you got time to post your solution? – Davide Giraudo Sep 29 '12 at 10:27

## 1 Answer

After looking into a few analysis books I found the strategy to make this work.

The idea is to use an analog of $G_\delta$ sets for an arbitrary measure space, and then show that the outer measure generated by $\mu$ on $R$ must be equal to $\mu$ for every set of finite outer measure.

More precisely, you show that any set $A \in \mathcal{A}$ can be approximated by an intersection of sets from $R$, meaning there exists $C_n\in R$ such that $A\subset \bigcap C_n$ and $\mu^*(\bigcap C_n \backslash A)=0$.

Then, since $\mu$ and $\mu^*$ agree on such intersections, it is not too hard to conclude that $\mu^*$ and $\mu$ must agree on every set $A \in \mathcal{A}$ with $\mu^*(A)<\infty$.

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Aren't your $C_n$s countable unions of sets in $R$, and so not necessarily in $R$ which is just an algebra? And you need to show that $\mu^*(A)<\infty$ when $\mu(A)<\infty$, but how do you do this? – Hamton Feb 11 '15 at 1:16