Proof of the change of variables formula without using the Monotone Convergence Theorem

lebesgue-integrallebesgue-measuremeasure-theoryreal-analysis

I recently encountered the problem Exercise 36 in Tao's An Introduction to Measure Theory. The link of an online version of this problem is here. Now I quote this problem as follows:

Exercise 36 (Change of variables formula)
Let $(X, \mathcal{B}, \mu)$ be a measure space, and let $\phi: X \rightarrow Y$ be a measurable morphism (as defined in Remark 8 from $(X, \mathcal{B})$ to another measurable space $(Y, \mathcal{C}). $ Define the pushforward $\phi_{*} \mu: \mathcal{C} \rightarrow[0,+\infty]$ of $\mu$ by $\phi$ by the formula $\phi_{*} \mu(E):=\mu\left(\phi^{-1}(E)\right)$

  1. Show that $\phi_{*} \mu$ is a measure on $\mathcal{C},$ so that $\left(Y, \mathcal{C}, \phi_{*} \mu\right)$ is a
    measure space.
  2. If $f: Y \rightarrow[0,+\infty]$ is measurable, show that $\int_{Y} f d \phi_{*} \mu=\int_{X}(f \circ \phi) d \mu$

(Hint: the quickest proof here is via the monotone convergence theorem below, but it is also possible to prove the exercise without this theorem.)

I really eager about how to prove the second statement WITHOUT the Monotone Convergence Theorem, in order to follow the procedure of the book. I tried hard and figured out only the case that $f$ is a simple function. How can we prove the case that $f$ is a general unsigned (nonnegative) function? The author have not provide the solution yet.

Any help is appreciated.

Best Answer

Recall that, by definition, $\int_Y f \ \mathrm{d}\phi_*\mu:=\sup\left\{\int_Y s \ \mathrm{d}\phi_*\mu : s \text{ is simple and } 0 \leq s \leq f\right\}$.

You already proved that $\int_Y s \ \mathrm{d}\phi_*\mu=\int_X (s\circ \phi) \ \mathrm{d}\mu$ for any simple function $s$. If $s$ is simple with $0\leq s \leq f$, then clearly $s \circ \phi$ is simple and $0 \leq (s \circ \phi)\leq (f \circ \phi)$. Thus, by the definition above we get

$$ \int_Y f \ \mathrm{d}\phi_*\mu \leq \int_X (f\circ \phi) \ \mathrm{d}\mu. $$

EDIT: For the reverse inequality, my first two arguments were flawed. I think that I've fixed it. See details below.

Let $t: X \to [0, +\infty]$ be a simple function with $t \leq (f \circ \phi)$, say given by $$ t:=\sum_{k=1}^n \alpha_k \chi_{E_k} $$ where $\alpha_k$ is such that $0 \leq\alpha_k \leq f(\phi(x))$ when $x \in E_k$. Define $$ F_k:=\{ y \in Y: f(y)\geq \alpha_k\}. $$ Then each $F_k$ is measurable because $f$ is measurable. Define a simple function $s: Y \to [0,+\infty]$ by $$ s:=\sum_{k=1}^n \alpha_k \chi_{F_k}. $$ By construction $0 \leq s \leq f$. Further, notice that $E_k=\{ x \in X: (f \circ \phi)(x) \geq \alpha_k\}$ and therefore it follows that $\phi^{-1}(F_k)=E_k$. Thus, by definition of the measure $\phi_*\mu$ we get $\int_X t \ \mathrm{d}\mu = \int_Y s \ \mathrm{d}\phi_*\mu$. Thus, taking sup over all simple functions $t \leq (f \circ \phi)$ we conclude that $$ \int_X (f\circ \phi) \ \mathrm{d}\mu \leq \int_Y f \ \mathrm{d}\phi_*\mu. $$ This is the reverse inequality.

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