Exercise 1.4.37 (Change of variables formula)

We first verify that ${\varphi }_{\text{∗}}\mu$ is a measure on $(Y,F)$.

(i)

We have ${\varphi }_{\text{∗}}\mu (\varnothing )=\mu ({\varphi }^{-1}(\varnothing ))=\mu (\varnothing )=0$.

(ii)

Let ${\left(E\right)}_{n\in \mathbb{N}}$ be a sequence of disjoint $F$-measurable sets. Notice that this implies that the sequence ${({\varphi }^{-1}(E_n))}_{n\in \mathbb{N}}$ is a sequence of disjoint $B$-measurable sets. We then have $${\varphi }_{\text{∗}}\mu \left(\bigcup _{n=1}^{\infty }{E}_n\right)=\mu \left({\varphi }^{-1}\left(\bigcup _{n=1}^{\infty }{E}_n\right)\right)=\mu \left(\bigcup _{n=1}^{\infty }{\varphi }^{-1}\left(E_n\right)\right)=\sum _{n=1}^{\infty }\mu \left({\varphi }^{-1}\left(E_n\right)\right)=\sum _{n=1}^{\infty }{\varphi }_{\ast }\mu \left(E_n\right).$$

We now demonstrate that

• Pick an arbitrary function $h$ minorising $f$, we then have

  $$\begin{array}{llll}\hfill \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{X}hd\mu & =\sum _{i=1}^k{a}_i\cdot {\varphi }_{\text{∗}}\mu \left(h^{-1}\left\{a_i\right\}\right)=\sum _{i=1}^k{a}_i\cdot \mu \left({\varphi }^{-1}\left(h^{-1}\left\{a_i\right\}\right)\right)& \hfill & \\ \hfill & =\sum _{i=1}^k{a}_i\cdot \mu \left({\left(h\circ \varphi \right)}^{-1}\left\{a_i\right\}\right)=:\sum _{i=1}^k{a}_i\cdot \mu \left({\left(g\right)}^{-1}\left\{a_i\right\}\right)& \hfill & \\ \hfill & =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{X}gd\mu & \hfill & \end{array}$$

  where we have defined $g:=h\circ \varphi$ on the fly. Since $h$ is simple, so is $h\circ \varphi$. Furthermore $h\le f$ necessarily implies that $h\circ \varphi \le f\circ \varphi$; thus, $g$ is a simple function minorising $f\circ \varphi$, and $\mathrm{Simp}{\int }_{X}gd\mu$ is contained in the set on the right-hand side.

• Pick an arbitrary function $g$ which is simple and minorises $f\circ \varphi$. We want to decompose our function $g$ into $h\circ \varphi$ for some $\varphi :X\to Y$ and a simple function $h:Y\to [0,+\infty ]$. To do so, denote

  $$\begin{array}{ccccc}\hfill X\hfill & \hfill \stackrel{\varphi }{\text{→}}\hfill & \hfill Y\hfill & \hfill \stackrel{h}{\text{→}}\hfill & \hfill [0,+\infty ]\hfill \\ \\ \hfill F_1:=g^{-1}\left(\left\{a_1\right\}\right)\hfill & \hfill \hfill & \hfill E_1:=\varphi (F_1)\hfill & \hfill \hfill & \hfill h(y):=a_1\text{ for }x\in E_1\hfill \\ \hfill ⋮\hfill & \hfill \hfill & \hfill ⋮\hfill & \hfill \hfill & \hfill ⋮\hfill \\ \hfill F_k:=g^{-1}\left(\left\{a_k\right\}\right)\hfill & \hfill \hfill & \hfill E_k:=\varphi (F_k)\hfill & \hfill \hfill & \hfill h(y):=a_k\text{ for }x\in E_k\hfill \\ \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill h(x):=0\text{ otherwise}\hfill \\ \hfill \hfill \end{array}$$

  Then we can easily verify that $g:=h\circ \varphi$. Notice that $h$ is simple, and for any $y\in Y$ we have $h(y)\le f(y)$. Thus, $\mathrm{Simp}{\int }_{X}hd\mu$ is contained in the set on the left-hand side. But we have

  $$\begin{array}{llll}\hfill \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{X}gd\mu & =\sum _{i=1}^k{a}_i\cdot \mu \left({\left(g\right)}^{-1}\left\{a_i\right\}\right)=\sum _{i=1}^k{a}_i\cdot \mu \left({\left(h\circ \varphi \right)}^{-1}\left\{a_i\right\}\right)& \hfill & \\ \hfill & =\sum _{i=1}^k{a}_i\cdot \mu \left({\varphi }^{-1}\left(h^{-1}\left\{a_i\right\}\right)\right)=\sum _{i=1}^k{a}_i\cdot {\varphi }_{\text{∗}}\mu \left(h^{-1}\left\{a_i\right\}\right)& \hfill & \\ \hfill & =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{X}hd\mu & \hfill & \end{array}$$