TITLE: Pullback measures
QUESTION [35 upvotes]: Why do all measure theory textbooks present the concept of push-forward measure, but never the concept of pull-back measure? Doesn't the latter exist?
It's true that the naive treatment of such a concept would sometimes lead to contradictions. For instance, let $p:\mathbb{R}^2\rightarrow\mathbb{R}$ be given by $p(x,y)=x$, $\lambda$ be the Lebesgue measure on $\mathbb{R}$, and $A=[0,1]\times[0,1]$. If one decomposes $A=A_1 \cup A_2$ where $A_1=[0,1]\times[0,\frac{1}{2}]$ and $A_2=[0,1]\times(\frac{1}{2},1]$, then $(p^* \lambda)(A)=1$ while $(p^* \lambda)(A_1)+(p^* \lambda)(A_2)=2$, showing that the naive definition of the pull-back does not lead to a measure.
Similarily, if $i:\mathbb{R}\rightarrow\mathbb{R}^2$ is given by $i(x)=(x,0)$, then the pull-back of the Lebesgue measure on $\mathbb{R}^2$ would be 0.
Given the two situations presented above, can one define a fruitful concept of pull-back measure?

REPLY [10 votes]: A measure $\mu$ on a set $X$ is a linear functional  from a  vector space of functions $F(X)$ on that set $X$ $\newcommand{\bR}{\mathbb{R}}$
$$ \mu : F(X)\to\bR,\;\; f\mapsto \langle \mu, f\rangle:=\int_X f(x) \mu(dx). $$
We can view the space of measures $M(X)$ on $X$  as a dual to $F(X)$. The natural  operation  on functions is that of pullback.  The natural operation on measures would be the dual operation, pushforward. Thus, if $\Phi: X\to Y$ is a map then we get two linear maps
$$ \Phi^*: F(Y)\to F(X),\;\;\Phi_\ast: M(X)\to M(Y) $$
related by the equality
$$ \langle \mu , \Phi^* f\rangle= \langle \Phi_* \mu, f\rangle,\;\;\forall f\in F(Y),\;\;\mu\in M(X). $$