Push-forward measure and change of variables

Let

$(\mathcal{X},A),$ and

$(\mathcal{Y},B)$ be measurable spaces, and

$(\alpha :A\to \mathbb{R})\in \mathcal{M}(\mathcal{X})$ be a measure on

$\mathcal{X}$. Let

$T:\mathcal{X}\to \mathcal{Y}$ be a measurable map.

The push-forward measure

$\beta =T_{\ast }\alpha$ of

$\alpha$ under

$T$ is defined by

$$\beta =\alpha \circ T^{-1}$$

Claim:

$\beta \in \mathcal{M}(\mathcal{Y})$, etc.

• In discrete case:
  $\alpha =\sum _i{a}_i{\delta }_{x_i},$
  $\beta =\sum _i{a}_i{\delta }_{T(x_i)}$.
• For a simple
  $g$ on
  $\mathcal{Y},$ say
  $g(x)=\sum _i{g}_i{I}_{x\in B_i}$, we would have
  $\int g\mathrm{d}\beta =\sum _i{g}_i\beta (B_i)=\sum _i{g}_i\alpha (T^{-1}(B_i)).$
• On the other hand
  $\int g\circ T\mathrm{d}\alpha =\sum _i{g}_i\alpha (\{T(x)\in B_i|x\in \mathcal{X}\})=\sum _i{g}_i\alpha (T^{-1}(B_i)).$
• TLDR:
  $$\int g\mathrm{d}(\alpha \circ T^{-1})=\int g\circ T\mathrm{d}\alpha .$$
• That's arithmetics and let's for now leave the real math (like, showing convergence and stuff) to real mathematicians. What I mean is, we further believe that this formula holds in general

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Densities

Let

$\alpha$ have density

$f_{\alpha }$ (w.r.t. some measure

$\lambda$):

$\alpha (f)={\int }_{\mathcal{X}}{f}_{\alpha }(x)f(x)\mathrm{d}\lambda x$. We want to find density of

$\beta$. So, there should be some measure in

$\mathcal{Y}$ to compute density with respect to.

Ok, let's think for now that

$\mathcal{Y}=\mathcal{X}$,

$A$ is Borel, and

$\lambda$ is the Lebesgue measure.

So,

$T:\mathcal{X}\to \mathcal{X}$.

$${\int }_{\mathcal{X}}{f}_{\alpha }(x)(f\circ T)(x)\mathrm{d}x={\int }_{\mathcal{Y}}f\mathrm{d}\beta ={\int }_{\mathcal{Y}}{f}_{\beta }f\mathrm{d}\lambda ,$$

$$f_{\beta }=?.$$

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Cartan

• $E,F$ – Banach spaces.
• $L_p(E;F)$ –
  $p$-multilinear forms,
  $E^p\to F$.
• $A_p(E;F)\subset L_p(E;F)$ – alternating
  $p$-linear forms.
• $U\subset E$ open.
• $\omega :U\to A_p(E;F)$ diff.
  $p$-form of class
  $C^n$.
• $E^{\prime }$ – Banach space.
  $U^{\prime }\subset E^{\prime }$ open.
• $\varphi :U^{\prime }\to U$ of class
  $C^{n+1}$.
• ${\varphi }^{\prime }$ derivative in the sense of Banach spaces/Frechet derivative.
• Thus
  $\varphi (y):E^{\prime }\stackrel{\sim }{\to }E.$
• Cartan denotes vectors
  ${\eta }_1,\dots ,{\eta }_p$, but let us use capital latins (
  $X_1,\dots ,X_p,Y_1,\dots$) instead.
• ${\omega }^{\prime }={\varphi }^{\ast }\omega :U^{\prime }\to A_p(E^{\prime };F)$.
• $y\in U^{\prime }.$
• $Y_1,\dots ,Y_p\in E^{\prime }$.
• $({\varphi }^{\ast }\omega )(y;Y_1,\dots ,Y_p)=\omega (\varphi (y);{\varphi }^{\prime }(y)Y_1,\dots ,{\varphi }^{\prime }(y)Y_p).$
• $({\varphi }^{\ast }\omega )(\underset{U^{\prime }}{\underbrace{y}};Y_1,\dots ,\underset{E^{\prime }}{\underbrace{Y_p}})=\omega (\underset{U}{\underbrace{\varphi (y)}};\underset{E}{\underbrace{{\varphi }^{\prime }(y)Y_1}},\dots ,{\varphi }^{\prime }(y)Y_p).$
• Cartan suggests we start with
  $p=0$ when
  $A_p(E;F)=F$ and
  $\omega =f$ is just a
  $C^n$ function
  $U\to F$.
• ${\varphi }^{\ast }\omega$ thus should be
  $U^{\prime }\to F.$
• $({\varphi }^{\ast }f)(y^{\prime })=f(\varphi (y^{\prime })).$
• $f\circ \varphi$ is still
  $C^n$ if
  $\varphi$ is
  $C^n$ too.
• Now
  $p>0$.
• TLDR: it's all compositions and forks of
  $C^n$ stuff, so it's
  $C^n$

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Yes, I'm stupid, ignorant, un-educated, absolutely hopeless moron. I just keep trying to fix that.