A problem concerning outer measure Is the following true for outer measure?
$\forall j\in\mathbb{N},A_j\subset \mathbb{R}^n$.Then 
$$m^*\left(\bigcup_{j\in \mathbb{N}}A_j\right)=\lim_{N\to\infty}m^*\left(\bigcup_{j=0}^{N}A_j\right)$$
 A: It's false in general for outer measures. I decided to dig this counter-example because there seemed to be an accepted misunderstanding in this topic.
This was found from Paul R. Halmos, Measure Theory, 1974, page 53, exercise (2). He states that it is false for non-regular outer measures in exercise (4), and gives this type of construction as a hint from exercise (2).
We say that $A\subset \mathbb{R}$ has infinite condensation if there are uncountably many points of $A$ outside of every bounded interval. Denote $2^{\mathbb{R}}$ as the powerset of $\mathbb{R}$.
Let $m^{*}:2^{\mathbb{R}}\to [0,\infty]$ be such that:


*

*$m^{*}(A)=0$ if $A$ is countable.

*$m^{*}(A)=1$ if $A$ is uncountable but does not have infinite condensation.

*$m^{*}(A)=\infty$ if $A$ has infinite condensation.
Then $m^{*}$ is an outer-measure and the only $m^{*}$-measurable sets are countable sets and sets with countable complements. You may check that for example $m^{*}([0,1])=1$ and that there exists no sets containing $[0,1]$ with countable complement and no infinite condensation. Hence $m^{*}$ is not regular.
Now let $A_{i}=[-i,i]$ for every $i\in \mathbb{N}$ whence $A_{1}\subset ... \subset A_{i}\subset A_{i+1}\subset ...\subset \mathbb{R}$. Each $A_{i}$ is uncountable and does not have infinite condensation so $m^{*}(A_{i})=1$ for each $i\in\mathbb{N}$. But $\mathbb{R}=\bigcup_{i=1}^{\infty}A_{i}$ and $m^{*}(\mathbb{R})=\infty$. Hence:
\begin{align*}\lim_{i\to\infty}m^{*}(A_{i})=1\neq \infty =m^{*}(\mathbb{R})=m^{*}(\bigcup_{i=1}^{\infty}A_{i})
\end{align*}
As required.
A: Yes, it is true for Lebesgue outer measure. The proof is from Zygmund & Wheeden Measure and Integral:

(3.26) is just the desired result for measurable sets.
