Pushforward/image outer measure Let $X$ and $Y$ be two sets, $f:X\to Y$ and $\mu$ be an outer measure on $2^X$. Is that true that the image of $\mu$ under $f$ defined by
$$
  \nu(B):= \mu(f^{-1}(B))
$$
is an outer measure on $Y$? If I am not mistaken, the answer is yes as $f^{-1}$ preserves empty sets, set inclusions and commutes with the unions. At the same time, I have never seen such construction in use, so I would be happy if anybody knows any reference on a topic.

The two features of an outer measure $\nu$ are 


*

*$\nu(\emptyset) = 0$, and

*$\displaystyle A \subset \bigcup_k B_k \implies \nu(A) \le \sum_k \nu(B_k)$


whenever $\{B_k\}$ is a countable collection of sets.
For the first part you have $$\nu(\emptyset) = \mu(f^{-1}(\emptyset)) = \mu(\emptyset) = 0$$
and for the second you have
\begin{align*}
A \subset \bigcup_k B_k &\implies f^{-1}(A) \subset f^{-1} \left( \bigcup_k B_k \right) = \bigcup_k f^{-1}(B_k) \\
&\implies \mu(f^{-1}(A)) \le \sum_k \mu(f^{-1}(B_k))
\end{align*}
since $\mu$ is an outer measure. Thus
$$
A \subset \bigcup_k B_k \implies \nu(A) \le \sum_k \nu(B_k)
$$
giving you that $\nu$ is an outer measure.
