I'm trying to strengthen my understanding of Bochner integral by extending change-of-variables formula to Banach spaces, i.e.,
Theorem: Let $(X, \mathcal X, \mu)$ be a $\sigma$-finite measure space, $(Y, \mathcal Y)$ a measurable space, $\pi:X \to Y$ measurable, and $(E, |\cdot|)$ a Banach space. Here we use Bochner integral. Let $\nu := \pi_\sharp \mu$ and $f \in L_0(Y, \nu, E)$. Then $f \in L_1 (Y, \nu, E)$ if and only if $f \circ \pi \in L_1(X, \mu, E)$. In this case, $$ \int_X (f \circ \pi) \mathrm d \mu = \int_Y f \mathrm d \nu \quad (\star). $$
Could you have a check on my attempt? Thank you so much!
Proof:
- $E := [0, \infty]$.
We have $f \in L_0 (Y, \nu, E)$ if and only if there is an increasing sequence $(f_n) \subset \mathcal S (Y, \nu, E)$ such that $f_n \to f$ $\nu$-a.e. It follows that $(f_n \circ \pi)$ is increasing and $f_n \circ \pi \to f \circ \pi$ $\mu$-a.e. By definition, $$ \int_X (f \circ \pi) \mathrm d \mu = \lim_n \int_X (f_n \circ \pi) \mathrm d \mu \quad \text{and} \quad \int_Y f \mathrm d \nu = \lim_n \int_Y f_n \mathrm d \nu. $$
Because $(\star)$ trivially holds for $\mu$-simple functions, we get $$ \int_X (f \circ \pi) \mathrm d \mu = \int_Y f \mathrm d \nu. $$
Our reasoning also shows that $f \in L_1 (Y, \nu, E) \iff f \circ \pi \in L_1(X, \mu, E)$.
- $E$ is a Banach space.
We have $f \in L_1 (Y, \nu, E)$ if and only if $|f| \in L_1 (Y, \nu, [0, \infty))$. Similarly, $f \circ \pi \in L_1(X, \mu, E)$ if and only if $|f| \circ \pi \in L_1(X, \mu, [0, \infty))$. We apply (1.) on $|f|$ to get the integrability equivalence between $f$ and $f \circ \pi$. Next we prove $(\star)$.
a. $f$ is $\nu$-simple.
Then $f = \sum_{i=1}^n e_i 1_{A_i}$ for some $e_i \in E\setminus \{0\}$ and $A_i \in \mathcal Y$ such that $(e_i)$ is pairwise different and $\nu(A_i) <\infty$. It follows that $f \circ \pi$ is $\mu$-simple. In fact, $f \circ g = \sum_{i=1}^n e_i 1_{B_i}$ with $B_i := \pi^{-1} (A_i)$. Notice that $\mu(B_i) = \mu(\pi^{-1} (A_i)) = \nu(A_i) < \infty$ It's clear that $(\star)$ holds.
b. $f \in L_1 (Y, \nu, E)$.
By definition, there is a Cauchy sequence $(f_n)$ of $\nu$-simple functions such that $f_n \to f$ $\nu$-a.e. As shown in (2.a.), $f_n \circ \pi$ is $\mu$-simple. It's clear that $f_n \circ \pi \to f \circ \pi$ $\mu$-a.e. Let's prove that $(f_n \circ \pi)$ is Cauchy. Indeed, $$ \begin{align} \|f_n \circ \pi - f_m \circ \pi\|_{L_1} = \|(f_n-f_m) \circ \pi\|_{L_1} = \int_X |(f_n-f_m) \circ \pi| \mathrm d \mu = \int_X |f_n-f_m| \circ \pi \mathrm d \mu. \end{align} $$
Notice that $|f_n-f_m|$ is $\nu$-simple. By (2.a.), $$ \int_X |f_n-f_m| \circ \pi \mathrm d \mu = \int_Y |f_n-f_m| \mathrm d \nu = \|f_n-f_m\|_{L_1}. $$
As such, $(f_n \circ \pi)$ is a Cauchy sequence. By definition, $$ \int_X (f \circ \pi) \mathrm d \mu = \lim_n \int_X (f_n \circ \pi) \mathrm d \mu \quad \text{and} \quad \int_Y f \mathrm d \nu = \lim_n \int_Y f_n \mathrm d \nu. $$
By (2.a.), we have $$ \int_X (f_n \circ \pi) \mathrm d \mu = \int_Y f_n \mathrm d \nu \quad \forall n. $$
Then $(\star)$ follows.
