# Outer measured induced by a point measure over an algebra of sets

I'm helping a friend with problem 7 from section 3.3 of Vestrup's 'Theory of Measures and Integrals'. We got partway into the problem, but after a few hours of chatting about it we were still stumped. I think it's a fun problem (though not exactly research-grade), so I'm posting it here!

First we fix a set $\Omega$, a point $\omega \in \Omega$, and an algebra of sets $A \subset 2^\Omega$ such that $\{\omega\}$ is an atom of $\sigma(A)$.

Then define $\mu(M) = 1_M(\omega)$ for $M \in A$. We define the outer-measure associated with $\mu$ by letting $\mu^*(M) = \inf \{\sum{ \mu(F_i)}\ |\ F_i \in A,\ M \subset \cup F_i \}$, where $F_i$ are at most countable $A$-covers of M.

The question is what is $\mu^*$? What are the measurable sets of $\mu^*$?

Some work so far:

If $\omega \in M$, then $\mu^*(M) = 1$. To see this, note that any $A$-cover of $M$ must cover $\omega$ at least once, and hence $\mu^*(M) \geq 1$. However $\mu(\Omega) = 1$ bounds things sharply from above.

On the other hand, if $M$ is nonempty and $\omega \notin M$, it seems reasonable that $\mu^*(M) = 0$. However, we had a hard time ruling out the possibility where $M$ does not admit any $\omega$-free $A$-covers. We think the condition on the generated $\sigma$-algebra of $A$ is key to understanding what happens, but don't know how to apply it.

If we relax the condition that $\{\omega\} \in \Sigma(A)$, we can get a very trivial outer measure. Let $A = \{\Omega, \emptyset\}$, and fix an arbitrary $\omega$. Then $\mu^*(M) = 1$ for any non-empty $M$, and is zero otherwise. In this case $\sigma(A) = A$ and $\{\omega\}$ is nowhere to be found.

On the other hand, let $\Omega = [0,1]$, fix $\omega = 0$, and let $A$ by the algebra generated by the family of sets $F_i = [1, 1/i]$. In this case, $\{0\}$ is not an atom of $A$, however it does belong to $\sigma(A)$. But now, we get things like $\mu^*((0,1]) = 1$ and in general, $\inf{M} = 0$ entails $\mu^*(M) = 1$, since all $A$-covers of $M$ must include $0$.

So, barring any mistakes, it seems like it may not be so easy to characterize $\mu^*$ !

[Edit:] It looks like I did make a mistake. Namely, in the example I gave, $A$ has the sets $G_i = (1/i, 1]$, being the complements of the $F_i$. But $G_i$ eventually covers any individual point in $(0,1]$, without ever including $0$ so it turns out $\mu*((0,1]) = 0$, as expected. I do not yet know how to show if $0 \in \sigma(A)$ is necessary or sufficient for this.

Is there a deeper story here?

• cads maybe the problem is wrong to begin with ? or can we solve it ?
– user111750
Feb 21, 2015 at 0:13
• Lol, Karim, at some point we'll bust it open! I'm almost sure that the answer has to be what we expect ($\mu^*(M) = 1_M(\omega)$ for any $M$). Suppose there is a downward chain of $A$-sets $F_1 \supset F_2 \supset F_3 ...$ with the property that $\omega \in F_i$ for all $i$, and for any $a \neq \omega$ we have some $k$ such that $a \notin F_k$. In that case, then clearly we have $\bigcap F_i = \{\omega\}$, hence $\{\omega\} \in \sigma(A)$ as required. Now suppose $A$ does not have such a chain. If we're lucky, proposing $\{\omega\} \in \sigma(A)$ now leads us to an easy contradiction. Feb 21, 2015 at 3:34
• I have added a bounty to this question I am really interested to know the answer..
– user111750
Feb 21, 2015 at 23:28
• We're not guaranteed $F_i - \{\omega\} \in A$, only that $\{\omega \}$ appears in the sigma algebra generated by $A$. By definition an algebra is closed under finite unions and intersections. The process of passing to closure under countable unions may introduce new singletons. Feb 22, 2015 at 4:06
• For example, let $A \subset 2^{[-1,1]}$ be the algebra generated by the intervals $[-1/n, 1/n]$ where $n \in {1, 2, ..}$. Then $\{0\} = \bigcap_1^\infty [-1/n, 1/n]$ is evidently in $\sigma(A)$. But if $F \in A$ and $0 \in F$, we fail to see $F - \{0\} \in A$. Feb 22, 2015 at 4:09

If $\{\omega\}\in\sigma(A)$, then $\mu^*(\Omega\backslash\{\omega\})=\mu^*(\Omega)-\mu^*(\{\omega\})=1-1=0$, since on $\mu^*$-measurable sets the outer measure is additive. It is also monotonous on all sets, so the outer measure of every set not containing $w$ is zero.
Edit: by the request in comments, elaborating some bits: The family of $\mu^*$-measurable sets is the $\sigma$-algebra obtained during the procedure of the Lebesgue continuation of $\mu$ from the algebra $A$ (it can be found in any textbook on the real/functional analysis). $\sigma(A)$ being the minimal $\sigma$-algebra containing $A$ thus consists of $\mu^*$-measurable sets.
• Thanks for the answer! Can you please expand on why $\{\omega\} \in \sigma(A)$ entails that $\{\omega\}$ is $\mu^*$-measurable? I feel like that's the crux of the difficulty we're having here. Feb 22, 2015 at 7:54
• The family of $\mu^*$-measurable sets constitute a $\sigma$-algebra, thus containing the minimal sigma-algebra generated by $A$, which I assume was meant by $\sigma(A)$. By "$\mu^*$-measurable" I mean the $\sigma$-algebra obtained as the Lebesgue continuation of $\mu$ from $A$, that is, the sets $T$ such that for every $\varepsilon$ $\mu^*(T\Delta S)<\varepsilon$ for some $S\in A$. Feb 22, 2015 at 8:26
• Thanks for the answer! After a little studying, it turns out it's fairly straight forward to see that the sigma algebra of $\mu^*$-measurable sets contains $A$. I've suggested an edit that references the appropriate proof. Cheers! Feb 22, 2015 at 11:03