Substitution formula §

Theorem Let $(X,\mathcal{F})$ and $(Y,\mathcal{E})$ be two measurable spaces, and a measurable map $h:X\to Y$, and a measurable function $g:Y\to \mathbb{R}$ (assuming the Borél $\sigma$-algebra). Then, if any of the two following integral exists, they are equal:

$${\int }_{Y}gd(h_{\ast }\mu )={\int }_{X}g\circ hd\mu$$

Proof Use the standard machinery. Assume $g=1_C$ with $C\in \mathcal{E}$ a measurable set. By definition of the lebesgue integral of a simple function

$${\int }_Y{1}_{C}d(h_{\ast }\mu )=h_{\ast }\mu (C)=\mu (h^{-1}(C))$$

by definition of the Pushforward measure. Now for the left side, using an integration variable $x$ for clarity

$${\int }_X{1}_C(h(x))d\mu (x)$$

the integrand function is

$$1_c(h(x))=\{\begin{array}{l}1\text{ se }h(x)\in C\\ 0\text{ se }h(x)\notin C\end{array}$$

this is the same as $1_{h^{-1}(C)}$ the characteristic function of the set $h^{-1}(C)$. This is again a simple function, so

$$=\mu (h^{-1}(C))$$

so the theorem is proved for characteristic functions of measurable set. By linearity, it follows easily for simple function $g={\sum }_{i=1}^n{k}_i{1}_{C_i}$.

Using Monotone Convergence Theorem we can extend the result to non-negative measruable functions, choose a sequence of simple functions such that ${\varphi }_n\uparrow g$.

$${\int }_{Y}gd(h_{\ast }\mu )=\underset{n\to \infty }{\lim }{\int }_Y{\varphi }_{n}d(h_{\ast }\mu )=\underset{n\to \infty }{\lim }{\int }_X{\varphi }_n(h(x))dx={\int }_{X}g\circ hd\mu$$

using the fact that ${\varphi }_n\circ h$ are simple functions and

$${\varphi }_n\circ h\uparrow g\circ h$$

Extending the result to not non-negative function is done as usual, splitting the function in the positive and negative part. $\square$

Variant using the Lebasgue definition Suppose now that $g$ is non-negative $g:Y\to [0,\infty )$, using the definition of the Lebasgue integral

$${\int }_{Y}gd(h_{\ast }\mu )=\sup \left\{{\int }_Y\tilde{S}d(h_{\ast }\mu )|\tilde{S}\le g,\tilde{S}\text{ simple function}\right\}$$

the inequality in the definition is $\forall y\in Y\tilde{S}(y)\le g(y)$. But we are working with the pushforward measure, so we can resctrict this inequality only in the image of $X$ under $h$:

$$\forall y\in h(X)\tilde{S}(y)\le g(y)$$

this is equivalent to

$$\forall x\in X\tilde{S}(h(x))\le g(h(x))$$

the inequality can be rewritten to

$$\tilde{S}\circ h\le g\circ h$$

Let’s get back to the definition of the integral above, and use the fact that we proved the substitution formula for the integral of simple functions

$$\sup \left\{{\int }_X\tilde{S}\circ hd\mu |\tilde{S}\circ h\le g\circ h,\tilde{S}\circ h\text{ simple function}\right\}$$

where $\tilde{S}:Y\to [0,\infty )$. Also the composition $\tilde{S}\circ h$ is a simple function but from $X\to [0,\infty )$. Just call $S:=\tilde{S}\circ h$

$$\sup \left\{{\int }_{X}Sd\mu |S\le g\circ h,S\text{ simple function}\right\}={\int }_{X}g\circ hd\mu$$

note that we are ignoring the restriction to simple function of the form $\tilde{S}\circ h$, removing this constraint doesn’t change the supremum.

Extending the result to not non-negative function is done as usual, splitting the function in the positive and negative part. $\square$