Question

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?

Answer 1

To define pullbacks of measures we need some additional data, because otherwise one would be able to obtain a canonical measure on an arbitrary measurable space M by pulling back the canonical measure on the point along the unique map M→pt.

One natural choice for such additional data is a choice of measure on each fiber of the map f: M→N. Using such a fiberwise measure one can define the pullback measure f*μ on M given a measure μ on N as follows: to integrate a function h on M with respect to f*μ we integrate h fiberwise with respect to the fiberwise measure on M and then we integrate the resulting function on N with respect to μ.

To define pullbacks of complex valued measures we have to require that the fiberwise measure (which can now also be complex valued) is fiberwise finite and its norm (total variation) is uniformly bounded with respect to N.

To ensure that pullbacks of probability measures are again probability measures we have to require that fiberwise measures are probability measures.

All of this can be done in greater generality for arbitrary noncommutative measurable spaces (i.e., von Neumann algebras) and arbitrary L_p-spaces (instead of just L_1-spaces, i.e., measures), where p is an arbitrary complex number, as explained in this answer: http://mathoverflow.net/questions/20740/is-there-an-introduction-to-probability-theory-from-a-structuralist-categorical-p/20820#20820

Answer 2

A simple-minded answer. The push forward of a measure is a triviality: take a measure space $(X,\mathcal M, \mu)$ and a mapping $f:X\rightarrow Y$. Then defining $\mathcal N=${$B\subset Y, f^{-1}(B)\in \mathcal M$}, you find easily that $\mathcal N$ is a $\sigma$-algebra on $Y$ and defining $$ \nu=f_*(\mu) \text{ measure on $(Y,\mathcal N)$},\quad \nu(B)=\mu(f^{-1}(B)), $$ you get that $(Y,\mathcal N,\nu)$ is a measure space (and $\mathcal N$ is the largest $\sigma$-algebra on $Y$ making $f$ measurable).

Now, the pullback: take a measure space $(X,\mathcal M, \mu)$ and a mapping $f:Z\rightarrow X$. You would like to find a measure space $(Z,\mathcal T, \omega)$ such that $f_*(\omega)=\mu$. It is possible when $f$ is bijective: in that case just take

$ \omega={(f^{-1})}_* $ $(\mu)= f^*(\mu).$ The last equality is a definition.

When $f$ is not bijective this program is unrealistic.

Answer 3

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 viu 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

$$ \langle \mu , \Phi^* f\rangle= \langle \Phi_* \mu, f\rangle,\;\;\forall f\in F(Y),\;\;\mu\in M(X). $$