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.
Best Answer
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".