# The Unapologetic Mathematician

## Pulling Back and Pushing Forward Structure

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 $(Y,\mathcal{T})$ is a measurable space and consider a function $f:X\to Y$. We can always make $f$ into a measurable function by pulling back the $\sigma$-ring $\mathcal{T}$. For each measurable subset $E\subseteq Y$ we define the preimage $f^{-1}(E)=\{x\in X\vert f(x)\in E\}$ as usual, and define the pullback $f^{-1}(\mathcal{T})$ to be the collection of subsets of $X$ of the form $f^{-1}(E)$ for $E\in\mathcal{T}$. Taking preimages commutes with arbitrary setwise unions and setwise differences, and $f^{-1}(\emptyset)=\emptyset$, and so $f^{-1}(\mathcal{T})$ is itself a $\sigma$-ring. Every point $x\in X$ gives us a point $f(x)\in Y$, and every point $f(x)\in Y$ is contained in some measurable set $E\in\mathcal{T}$. Thus $x$ is contained in the set $f^{-1}(E)\in f^{-1}(\mathcal{T})$, and so we find that $(X,f^{-1}(\mathcal{T}))$ is a measurable space. Clearly, $f^{-1}(\mathcal{T})$ contains the preimage of every measurable set $E\in\mathcal{T}$, and so $f:(X,f^{-1}(\mathcal{T}))\to(Y,\mathcal{T})$ is measurable.

Measures, on the other hand, go the other way. Say that $(X,\mathcal{S},\mu)$ is a measure space and $f:(X,\mathcal{S})\to(Y,\mathcal{T})$ is a measurable function between measurable spaces, then we can define a new measure $\nu$ on $Y$ by “pushing forward” the measure $\mu$. Given a measurable set $E\subseteq Y$, we know that its preimage $f^{-1}(E)\subseteq X$ is also measurable, and so we can define $\nu(E)=\mu(f^{-1}(E))$. It should be clear that this satisfies the definition of a measure. We’ll write $\nu=f(\mu)$ for this measure.

If $f:X\to Y$ is a measurable function, and if $\mu$ is a measure on $X$, then we have the equality

$\displaystyle\int g\,d(f(\mu))=\int(g\circ f)\,d\mu$

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 $g=\chi_E$ for a measurable set $E\subseteq Y$. 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 $\chi_E$ is the characteristic function of $E$, then $\left[\chi_E\circ f\right](x)=1$ if $f(x)\in E$ — that is, if $x\in f^{-1}(E)$ — and $0$ otherwise. That is, $\chi_E\circ f=\chi_{f^{-1}(E)}$. We can then calculate

$\displaystyle\int\chi_E\,d(f(\mu))=\left[f(\mu)\right](E)=\mu(f^{-1}(E))=\int\chi_{f^{-1}(E)}\,d\mu=\int(\chi_E\circ f)\,d\mu$

As a particular case, applying the previous result to the function $g\chi_E$ shows us that

\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 $y=f(x)$.

Finally, we can combine this with the Radon-Nikodym theorem. If $f:X\to Y$ is a measurable function from a measure space $(X,\mathcal{S},\mu)$ to a totally $\sigma$-finite measure space $(Y,\mathcal{T},\nu)$ so that the pushed-forward measure $f(\mu)$ is absolutely continuous with respect to $\nu$. Then we can select a non-negative measurable function

$\displaystyle\phi=\frac{d(f(\mu)}{d\nu}:Y\to\mathbb{R}$

so that

$\displaystyle\int g(f(x))\,d\mu(x)=\int g(y)\phi(y)\,d\nu(y)$

again, in the sense that if one of these integrals exists then so does the other, and their values are equal. The function $\phi$ plays the role of the absolute value of the Jacobian determinant.

August 2, 2010 - Posted by | Analysis, Measure Theory

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