Let $X\neq \emptyset$ and $\mu^*\colon \mathcal{P}(X)\to\left[0,\infty\right]$ be an outer measure. Let $(A_i)_{i=1}^\infty$ be a sequence of sets in $X$ such that $A_i\subset A_{i+1}$ for all $i\in \mathbb{N}$. Show that, without assuming that the sets $A_i$ are $\mu^*$-measurable,
$$ \mu^*\left(\bigcup_{i=1}^\infty A_i\right)=\lim_{i\to\infty} \mu^*(A_i) $$
is not true in general.
My attempt
Since a counterexample cannot be found with the Lebesgue outer measure on $\mathbb{R}^n$, I tried concocting some other counterexample. Here's what I came up with:
Let $X=\mathbb{R}$ and $\nu^*$ be the counting measure on $X$, i.e. $\nu^*(A)=\#A$ if $A$ is finite and $\nu^*(A)=\infty$ otherwise. Then define
$$ \mu^*(A)=\begin{cases} \tanh \nu^*(A), &A \ \text{is finite} \\ 2, &A \ \text{is infinite} \end{cases} $$
for all $A\subset X$. Clearly $\mu^*(\emptyset)=\tanh\nu^*(\emptyset)=\tanh(0)=0$. Now let $A,B\subset X$ such that $A\subset B$. If $B$ is finite, then so is $A$ and
$$ \mu^*(A)=\tanh \nu^*(A)\leq \tanh \nu^*(B)=\mu^*(B), $$
since $\mu^*$ is an outer measure and $\tanh$ is an increasing function. If $B$ is infinite and $A$ is finite,
$$ \mu^*(A)=\tanh \nu^*(A)\leq 1\leq 2=\nu^*(B). $$
Finally, if both $A$ and $B$ are infinite, then
$$ \mu^*(A)=2\leq 2=\mu^*(B). $$
Next, let $(A_i)_{i=1}^\infty$ be a collection of sets in $X$. If $\bigcup_{i=1}^\infty A_i$ is finite, this yields
$$ \mu^*\left(\bigcup_{i=1}^\infty A_i \right)=\tanh \nu^*\left(\bigcup_{i=1}^\infty A_i \right)\leq \tanh \left(\sum_{i=1}^\infty \nu^*(A_i) \right) $$
as $\tanh$ is increasing, and where $\tanh(\infty)=1$. Now, the addition formula for $\tanh$ says that
$$ \tanh(x+y)=\frac{\tanh x+\tanh y}{1+\tanh x\tanh y}. $$
Since $\tanh x\geq 0$ for all $x\geq 0$, we get
$$ \tanh(x+y)\leq \tanh x+\tanh y $$
for $x,y\geq 0$. This can be generalized inductively to
$$ \tanh\left(\sum_{i=1}^n x_i\right)\leq \sum_{i=1}^n \tanh x_i $$
for $x_i\geq 0$. Since $\tanh$ is continuous,
\begin{align*} \tanh \left(\sum_{i=1}^\infty \nu^*(A_i) \right)&=\tanh\left(\lim_{n\to\infty} \sum_{i=1}^n \nu^*(A_i) \right)=\lim_{n\to\infty} \tanh\left(\sum_{i=1}^n \nu^*(A_i) \right) \\ &\leq \lim_{n\to\infty} \sum_{i=1}^n \tanh \nu^*(A_i)=\sum_{i=1}^\infty \tanh \nu^*(A_i)=\sum_{i=1}^\infty \mu^*(A_i), \end{align*}
as $\nu^*(A_i)$ is always non-negative. If $\bigcup_{i=1}^\infty A_i$ is infinite, then there must be an infinite amount of non-empty sets. Let $I\in \mathbb{N}$ be the index set such that $i\in I$ if and only if $A_i\neq \emptyset$. Accordingly,
$$ \mu^*\left(\bigcup_{i=1}^\infty A_i \right)=2\leq \infty=\sum_{i\in I} \tanh(1)\leq \sum_{i\in I} \tanh \# A_i=\sum_{i\in I} \tanh \nu^*(A_i)=\sum_{i\in I}\mu^*(A_i)\leq \sum_{i=1}^\infty \mu^*(A_i). $$
Thus $\mu^*$ is an outer measure. Now, let $A_i=\{1,\ldots,i\}$ for all $i\in \mathbb{N}$. Then clearly
$$ \mu^*\left(\bigcup_{i=1}^\infty A_i \right)=\mu^*(\mathbb{N})=2. $$
However,
$$ \lim_{i\to\infty} \mu^*(A_i)=\lim_{i\to\infty} \tanh \nu^*(A)=\tanh \lim_{i\to\infty} \# A_i=\tanh(\infty)=1\neq 2=\mu^*\left(\bigcup_{i=1}^\infty A_i \right). $$
My questions
- Is my counterexample valid (i.e. is the proof correct)?
- Is there a simpler counterexample? (I'm sure there are many, but what are they?)