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The problem given to me on my homework is:

Prove that the limit of a decreasing family of outer measures is an outer measure.

Doing out the "obvious" approach, we quickly reach the problem of wanting to say that $$\lim_{j\to\infty}\;\mu_j^*\left(\bigcup_{i=1}^\infty A_i\right)\leq\lim_{j\to\infty}\left(\sum_{i=1}^\infty\;\mu_j^*(A_i)\right)\underbrace{\leq}_{\text{PROBLEM}}\sum_{i=1}^\infty\left(\lim_{j\to\infty}\;\mu_j^*(A_i)\right)$$ The problem, as I see it, is that Fatou's Lemma has the inequality going the opposite way of what we want here (by the way, this course hasn't actually gotten to integration yet).

In fact, I think the claim is false, because I can imagine very well having the following scenario:

  • $\{A_i\}_{i=1}^\infty\subset \mathcal{P}(X)$ is a family of subsets of $X$ (disjoint, probably)
  • $\mu_j^*:\mathcal{P}(X)\to[0,\infty]$ are decreasing family of outer measures on $X$ with the property that $$\mu_j^*(A_i)=\begin{cases}1\text{ if }j\leq i\\ 0\text{ if }j>i \end{cases}$$ (This certainly would not be in conflict with the assumption that the $\mu_j^*$ are decreasing, i.e. that $\mu_j^*(A)\geq\mu_{j+1}^*(A)$ for all $j\in\mathbb{N}$ and $A\subseteq X$.)

Letting $\mu^*:\mathcal{P}(X)\to[0,\infty]$ be defined by $\mu^*(A)=\lim_{j\to\infty}\mu_j^*(A)$, we conclude immediately that $\mu^*(\varnothing)=0$ and that $\mu^*(A)\leq\mu^*(B)$ when $A\subseteq B$, but we then have that $$\mu^*\left(\bigcup_{i=1}^\infty A_i\right)=\lim_{j\to\infty}\;\mu_j^*\left(\bigcup_{i=1}^\infty A_i\right)\geq\lim_{j\to\infty}\;\mu_j^*(A_j)=\lim_{j\to\infty}\;1=1$$ while $$\sum_{i=1}^\infty\;\mu^*(A_i)=\sum_{i=1}^\infty\left(\lim_{j\to\infty}\;\mu_j^*(A_i)\right)=\sum_{i=1}^\infty\left(\lim_{j\to\infty}\;\;{1\text{ if }j\leq i\atop 0\text{ if }j>i}\right)=\sum_{i=1}^\infty\;0=0 $$ so that $$\mu^*\left(\bigcup_{i=1}^\infty A_i\right)\not\leq\sum_{i=1}^\infty\;\mu^*(A_i)$$ and therefore $\mu^*$ is not an outer measure.

So, giving as little away as possible (as this is a homework question), is the scenario I proposed above impossible? I tried to construct examples with $X=\mathbb{N}$ and the value of $\mu_j^*(A)$ depending on whether $A\cap\{1,\ldots,j\}=\varnothing$, but that didn't get anywhere.

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Zev, I think I see it now. We define an outer measure on a set $X$ to be a function $\mu*:\mathcal{P}(X)\to[0,\infty]$, as you have mentioned. But in your counter-example, you assumed that the $A_i$ were disjoint to preserve countable additivity (at least, when I first heard the question, I thought that their disjoint-ness allowed us to maintain countable additivity.). I am almost certain that this is not, in fact, an outer measure now, because I do not think the sets can be enumerated (if they can be) in such a way to preserve additivity despite the fact that they are not actually disjoint. –  mixedmath Sep 25 '11 at 19:05
    
@mixedmath: But outer measures need only be countably subadditive; I was only proposing the $A_i$ be disjoint because it seems like it'd be easier to construct a counterexample when that's the case. (By the way, the original counterexample I'd proposed failed because the $\mu_j^*$ weren't monotonic; taking a set disjoint from $\{1,\ldots,j\}$ and adding in something from $\{1,\ldots,j\}$ decreased its $\mu_j^*$-measure.) –  Zev Chonoles Sep 25 '11 at 19:07
    
Aha, of course. How quick of me. And here, I was so pleased that it was suddenly so obvious. Hmm. –  mixedmath Sep 25 '11 at 19:16
    
Well, the exchange of limit and sum would be okay for an increasing sequence of measures anyway... –  Jesse Madnick Sep 25 '11 at 20:47
1  
Here's a positive variant of the result you ask about: 1. Let $\varphi: \mathcal{P}(X) \to [0,\infty]$ be any function such that $\varphi(\emptyset) = 0$. Put $$\mu_{\varphi}^{\ast}(A) = \inf{\left\{\sum_{n=0}^{\infty}\varphi(A_n)\,:\,A \subset \bigcup_{n=0}^{\infty} A_n\right\}}$$ then $\mu_{\varphi}^{\ast}$ is an outer measure. 2. If $\{\mu_{j}^{\ast}\}_{j \in J}$ is an arbitrary family of outer measures, let $\varphi(A) = \inf_{j \in J} \mu_{j}^{\ast}(A)$. Then $\mu_{\varphi}^{\ast}$ is the largest outer measure $\mu^{\ast}$ satisfying $\mu^{\ast} \leq \mu_{j}^{\ast}$. –  commenter Sep 26 '11 at 10:01

1 Answer 1

up vote 3 down vote accepted

Let $\gamma$ be the counting measure on the positive integers. Define $\mu^*_j(A) = \gamma( \{ x \in A | x \ge j \})$. That this sequence is decreasing follows from the monotonicity of $\gamma$. Since every finite set is bounded, its measure converges to 0 as $j \to \infty$. On the other hand, the measure of infinite sets is infinite for all $j$. This contradicts countable subadditivity of the limit.

I have not checked, but it seems likely that the exercise can be fixed by substituting "finite outer measure" for "outer measure".

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