Remember that we defined measurable functions in terms of inverse images, like we did for topological spaces. So it should be no surprise that we move a lot of measurable structure around between spaces by “pulling back” or “pushing forward”.
First of all, let’s say that 
is a measurable space and consider a function 
. We can always make 
into a measurable function by pulling back the 
-ring 
. For each measurable subset 
we define the preimage 
as usual, and define the pullback 
to be the collection of subsets of 
of the form 
for 
. Taking preimages commutes with arbitrary setwise unions and setwise differences, and 
, and so is itself a -ring. Every point 
gives us a point 
, and every point is contained in some measurable set . Thus 
is contained in the set 
, and so we find that 
is a measurable space. Clearly, contains the preimage of every measurable set , and so 
is measurable.
Measures, on the other hand, go the other way. Say that 
is a measure space and 
is a measurable function between measurable spaces, then we can define a new measure 
on 
by “pushing forward” the measure 
. Given a measurable set , we know that its preimage 
is also measurable, and so we can define 
. It should be clear that this satisfies the definition of a measure. We’ll write 
for this measure.
If is a measurable function, and if is a measure on , then we have the equality
in the sense that if either integral exists, then the other one does too, and their values are equal. As usual, it is sufficient to prove this for the case of 
for a measurable set . Linear combinations will extend it to simple functions, the monotone convergence theorem extends to non-negative measurable functions, and general functions can be decomposed into positive and negative parts.
Now, if 
is the characteristic function of 
, then =1](https://s0.wp.com/latex.php?latex=%5Cleft%5B%5Cchi_E%5Ccirc+f%5Cright%5D%28x%29%3D1&bg=e6e6e6&fg=333333&s=0 "\left[\chi_E\circ f\right](x)=1")
if 
— that is, if 
— and 
otherwise. That is, 
. We can then calculate
=\mu(f^{-1}(E))=\int\chi_{f^{-1}(E)}\,d\mu=\int(\chi_E\circ f)\,d\mu](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cint%5Cchi_E%5C%2Cd%28f%28%5Cmu%29%29%3D%5Cleft%5Bf%28%5Cmu%29%5Cright%5D%28E%29%3D%5Cmu%28f%5E%7B-1%7D%28E%29%29%3D%5Cint%5Cchi_%7Bf%5E%7B-1%7D%28E%29%7D%5C%2Cd%5Cmu%3D%5Cint%28%5Cchi_E%5Ccirc+f%29%5C%2Cd%5Cmu&bg=e6e6e6&fg=333333&s=0 "\displaystyle\int\chi_E\,d(f(\mu))=\left[f(\mu)\right](E)=\mu(f^{-1}(E))=\int\chi_{f^{-1}(E)}\,d\mu=\int(\chi_E\circ f)\,d\mu")
As a particular case, applying the previous result to the function 
shows us that
&=\int\limits_Eg\,d(f(\mu))\\&=\int g\chi_E\,d(f(\mu))\\&=\int(g\circ f)(\chi_E\circ f)\,d\mu\\&=\int(g\circ f)\chi_{f^{-1}(E)}\,d\mu\\&=\int\limits_{f^{-1}(E)}(g\circ f)\,d\mu\\&=\int\limits_{f^{-1}(E)}g(f(x))\,d\mu(x)\end{aligned}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7D%5Cint%5Climits_Eg%28y%29%5C%2Cd%5Cleft%5Bf%28%5Cmu%29%5Cright%5D%28y%29%26%3D%5Cint%5Climits_Eg%5C%2Cd%28f%28%5Cmu%29%29%5C%5C%26%3D%5Cint+g%5Cchi_E%5C%2Cd%28f%28%5Cmu%29%29%5C%5C%26%3D%5Cint%28g%5Ccirc+f%29%28%5Cchi_E%5Ccirc+f%29%5C%2Cd%5Cmu%5C%5C%26%3D%5Cint%28g%5Ccirc+f%29%5Cchi_%7Bf%5E%7B-1%7D%28E%29%7D%5C%2Cd%5Cmu%5C%5C%26%3D%5Cint%5Climits_%7Bf%5E%7B-1%7D%28E%29%7D%28g%5Ccirc+f%29%5C%2Cd%5Cmu%5C%5C%26%3D%5Cint%5Climits_%7Bf%5E%7B-1%7D%28E%29%7Dg%28f%28x%29%29%5C%2Cd%5Cmu%28x%29%5Cend%7Baligned%7D&bg=e6e6e6&fg=333333&s=0 "\displaystyle\begin{aligned}\int\limits_Eg(y)\,d\left[f(\mu)\right](y)&=\int\limits_Eg\,d(f(\mu))\\&=\int g\chi_E\,d(f(\mu))\\&=\int(g\circ f)(\chi_E\circ f)\,d\mu\\&=\int(g\circ f)\chi_{f^{-1}(E)}\,d\mu\\&=\int\limits_{f^{-1}(E)}(g\circ f)\,d\mu\\&=\int\limits_{f^{-1}(E)}g(f(x))\,d\mu(x)\end{aligned}")
We can go back and forth between either side of this equation by the formal substitution 
.
Finally, we can combine this with the Radon-Nikodym theorem. If is a measurable function from a measure space to a totally -finite measure space 
so that the pushed-forward measure 
is absolutely continuous with respect to . Then we can select a non-negative measurable function
so that
again, in the sense that if one of these integrals exists then so does the other, and their values are equal. The function 
plays the role of the absolute value of the Jacobian determinant.
August 2, 2010
Posted by John Armstrong |
Analysis, Measure Theory |
4 Comments
