TITLE: Outer measure discontinuous from below
QUESTION [6 upvotes]: I was trying to find an example of an outer Measure which is not continuous from below. 
These are the definitions I use

An outer measure on $X$ is a function $\mu^\ast: \mathcal{P}(X)\to [0,\infty]$ if 
  it fulfills

$\mu^\ast(\emptyset)=0$
$\mu^\ast\Big( \bigcup_{j=1}^\infty A_j\Big) \leq \sum_{j=1}^\infty A_j$


And an outer measure is continuous from below when for the sequence
$(A_j)_{j\in \mathbb{N}}$ with  $A_j\subset A_{j+1}$ for alle $j$ 
the equality 
$$ \mu^\ast \Big( \bigcup_{j=1}^\infty A_j\Big)= \lim_{j\to \infty} \mu^\ast (A_j)$$
Some results which might be helpful


All measures are continuous from below
All metric outer measures are continuous from below


So I search for an outer measure which isn't continuous from below.

REPLY [5 votes]: Let
$$\mu^\ast(A) = \begin{cases}
0\quad\;,\; A = \varnothing\\
1\quad\;,\; A \text{ is finite and nonempty}\\
\infty\quad, \text{ otherwise}\end{cases}$$
on an infinite set $X$.
If $\bigcup A_j$ is infinite, either at least one $A_j$ is infinite, or infinitely many $A_j$ are nonempty, so
$$\mu^\ast\left(\bigcup A_j\right) \leqslant \sum \mu^\ast(A_j).$$
Let $A_j = \{x_k\colon 1 \leqslant k \leqslant j\}$ for a sequence of distinct $x_k$, then
$$\mu^\ast(A_j) = 1$$
for all $j$, but $\mu^\ast\left(\bigcup A_j\right) = \infty$.