---
breaks: False
tags: [integration, measures, differential forms]
---

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 $\beta = T_*\alpha$
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 $\int g\operatorname{d}\beta = \sum_i g_i \beta(B_i) = \sum_i g_i \alpha(T^{-1}(B_i)).$
- On the other hand $\int g\circ T \operatorname{d}\alpha = \sum_i g_i \alpha(\{T(x)\in B_i\left|x\in \mathcal{X}\right.\}) = \sum_i g_i \alpha(T^{-1}(B_i)).$
- TLDR: $$\int g \operatorname{d}(\alpha\circ T^{-1}) = \int g\circ T \operatorname{d}\alpha.$$
- 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 $\lambda$): $\alpha(f) = \int_{\mathcal{X}} f_\alpha(x) f(x) \operatorname{d}\lambda x$.
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, $T: \mathcal{X}\to \mathcal{X}$.

$$\int_\mathcal{X} f_\alpha(x) (f\circ T)(x)\operatorname{d}x = \int_{\mathcal{Y}} f\operatorname{d}\beta = \int_\mathcal{Y} f_\beta f \operatorname{d}\lambda,$$
$$f_\beta =~?.$$

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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 $C^n$.
- $E'$ -- Banach space. $U'\subset E'$ open.
- $\phi: U' \to U$ of class $C^{n+1}$.
- $\phi'$ derivative in the sense of Banach spaces/Frechet derivative.
- Thus $\phi(y): E' \xrightarrow\sim E.$
- Cartan denotes vectors $\eta_1, \ldots, \eta_p$, but let us use capital latins ($X_1, \ldots, X_p, Y_1, \ldots$) instead.
- $\omega' = \phi^*\omega : U' \to A_p(E'; F)$.
- $y\in U'.$
- $Y_1, \ldots, Y_p \in E'$.
- $(\phi^*\omega)(y; Y_1, \ldots, Y_p) = \omega(\phi(y); \phi'(y)Y_1, \ldots, \phi'(y)Y_p).$
- $(\phi^*\omega)(\underbrace{y}_{U'}; Y_1, \ldots, \underbrace{Y_p}_{E'}) = \omega(\underbrace{\phi(y)}_{U}; \underbrace{\phi'(y)Y_1}_{E}, \ldots, \phi'(y)Y_p).$
- Cartan suggests we start with $p=0$ when $A_p(E; F)=F$ and $\omega=f$ is just a $C^n$ function $U\to F$.
- $\phi^*\omega$ thus should be $U' \to F.$
- $(\phi^*f)(y') = f(\phi(y')).$
- $f\circ \phi$ is still $C^n$ if $\phi$ is $C^n$ too.
- Now $p>0$.
- TLDR: it's all compositions and forks of $C^n$ stuff, so it's $C^n$


* * *

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