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?

  • 2
    $\begingroup$ See Tien-Cuong Dinh, Nessim Sibony, Pull-back of currents by holomorphic maps, manuscripta mathematica July 2007, Volume 123, Issue 3, pp 357–371 link.springer.com/article/10.1007/s00229-007-0103-5 $\endgroup$ – user21574 Oct 24 '17 at 21:55
  • $\begingroup$ Attention: If $M$ and $N$ be complex manifolds of the same dimension and $\pi : M\to N$ is a holomorphic mapping, then for a volume form (as measure )$\Psi$ on $N$ the pull-back $\pi^*\Psi$ is positive outside aramification divisor of $M$ and may not be a positiveon the whole of $M$. There is a classical paper of P. Griffiths (When I was master student in Marseille Ihad a course on Nevanlinna theory publications.ias.edu/sites/default/files/nevanlinna.pdf) $\endgroup$ – user21574 Oct 24 '17 at 22:07

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: Is there an introduction to probability theory from a structuralist/categorical perspective?

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    $\begingroup$ Upvote for the first sentence. $\endgroup$ – Adam Epstein Feb 23 '13 at 16:11
  • $\begingroup$ MO prevents me from doing such a small edit, but you probably wanted to write f\*μ instead of f*μ, otherwise the Markdown parser gets confused. $\endgroup$ – Alex Shpilkin Jan 31 '19 at 18:23
  • $\begingroup$ @AlexShpilkin: I wanted to write exactly what I wrote. Did you think that the answer would stay here for 6 years with such a glaring problem unnoticed? The Markdown parser was altered after the answer was written, which created a problem out of nowhere. I really wish the SE people would stop these pointless changes. I adjusted the syntax, it should work now. $\endgroup$ – Dmitri Pavlov Jan 31 '19 at 19:52
  • $\begingroup$ @DmitriPavlov Sorry, I could have phrased that better. As for changes in the parser... Well, Markdown is remarkably tolerant of unescaped markup (which is why your original text worked) on one hand, but extremely badly specified on the other (the original spec was literally “whatever this soup of Perl, PCREs and bugs produces”). So while a spherical programmer in a vacuum could probably carve their Markdown code in stone, a real-world implementation is bound to break from time to time. [cont.] $\endgroup$ – Alex Shpilkin Jan 31 '19 at 22:48
  • $\begingroup$ [cont.] Maintaining bug-compatible (actual term) implementations of garbage specs is a job noöne wants, and more importantly, noöne can reliably do. So I wouldn’t say the SE maintainers are at fault here; the original spec is garbage, and a passable one isn’t even done yet. Until it is, both users and maintainers are going to pay the price of the bad engineering choice made at the very beginning. [cont.] $\endgroup$ – Alex Shpilkin Jan 31 '19 at 23:04

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.

  • $\begingroup$ Measurability of $f^{-1}$ is needed for $f^\ast \mu$ to be a measure. $\endgroup$ – Arrow Dec 28 '16 at 12:07

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

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

  • $\begingroup$ Which of the two $\Phi$ is the pull back, vs. the pushforward? As someone who has studied Functional analysis and is currently studying Measure Theory, this post is helping me understand, but I need a bit more explanation about the dual spaces. Is the push forward and pull back dual to each other in the sense of Hilbert spaces? $\endgroup$ – Joshua Chen Oct 24 '17 at 17:06
  • 1
    $\begingroup$ The pullback is $\Phi^*$ and acts on spaces of functions via the equality $\Phi^*(f):=f\circ \Phi$. It is a contravariant functor hence the upper $*$. $\Phi_*$ is the dual of $\Phi^*$ in the sense of duality of locally convex topological spaces. The space of Radon measures on a locally compact space $X$ is the topological dual of the space of compactly supported continuous functions on $X$. This is not a Hilbert space. $\endgroup$ – Liviu Nicolaescu Oct 24 '17 at 23:56

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