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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.2011-09-25
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    @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.)2011-09-25
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    Aha, of course. How quick of me. And here, I was so pleased that it was suddenly so obvious. Hmm.2011-09-25
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    Well, the exchange of limit and sum would be okay for an [increasing sequence](http://en.wikipedia.org/wiki/Monotone_convergence_theorem#Convergence_of_a_monotone_series) of measures anyway...2011-09-25
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    If you take $\mu^*_j$ to be equal to zero measure on the integers up to $j$ and equal to counting measure elsewhere, I think that's your counterexample. On the other hand, could it be that the outer measures are meant to be finite?2011-09-25
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    @Niels: Thanks, that works! I'll bring it up to the teacher. If you'd like to post that as an answer, I'll accept it, or I can post my write-up of it as a Community Wiki answer.2011-09-26
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    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 $\mu^{\ast} \leq \mu_{j}^{\ast}$.2011-09-26

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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".