Proving inequality between a Lebesgue Integrals I have two non-zero real constants $a,b$ and a function $\gamma: \mathbb{R^2} \to \mathbb{R^2}$ where $\gamma(x,y) = (ax, by)$. I also have a integrable simple function $\phi: \mathbb{R^2} \to [0, \infty]$ and a Lebesgue-measurable function $r: \mathbb{R^2} \to [0, \infty]$ such that $0 \le \phi(x,y) \le r(x,y)$.
How can I show that $\int_\mathbb{R^2} \phi(x,y)d\mu \le |a||b|\int_{\mathbb{R^2}}r(\gamma(x,y))d\mu$?
I have tried a few approaches but none that lead very far. I know that $\mu(\gamma(E)) = |a||b|\mu(E)$ $\forall E\in \mathbb{R^2}$ where $E$ is Lebesgue measurable. I tried converting the simple function's integral to its canonical form, but that doesn't seem to lead me anywhere. I also tried to say that, since $\phi \leq r$, that $\int_\mathbb{R^2} \phi(x,y)d\mu \leq \int_{\mathbb R^2}r(x,y)d\mu$, but once again, I don't know how to move on from there. I feel like the question is very simple but that I'm missing some key idea. Any insight would be appreciated. Thanks.
 A: We shall prove this fact for $r$ a characteristic function satisfying the property.
Let us have $r=1_A:\mathbb R^2\to [0,\infty)$ (for some measurable set $A\subset \mathbb R^2$ ) defined by
$$
r(x,y)=1_A(x,y) =
\left\{
 \begin{array}{ll}
  1 & \mbox{if } (x,y) \in A \\
  0 & \mbox{otherwise } 
 \end{array}
\right.
$$
and we also have $0 \le \phi(x,y) \le r(x,y)=1_A(x,y)$.
So naturally we have

$$\int_\mathbb{R^2} \phi(x,y)d\mu \le \int_{\mathbb{R^2}}r(x,y)d\mu=\int_{\mathbb R^2}1_A(x,y)d\mu=\mu(A).$$

As we have $a,b\ne 0$ we have $\gamma$ is a bijective map implies $\gamma^{-1}:\mathbb R^2\to \mathbb R^2$ exist where $\gamma^{-1}(x,y)=\left(\frac{x}{a},\frac{y}{b}\right).$
Now look at
$$r(\gamma(x,y))=1_A(\gamma(x,y))=1_A(ax,by)=
\left\{
 \begin{array}{ll}
  1 & \mbox{if } (ax,by) \in A \implies (x,y)\in \gamma^{-1}(A)\\
  0 & \mbox{otherwise } 
 \end{array}
\right.\\ \\ \\$$$$
\therefore 1_A(\gamma(x,y))=1_{\gamma^{-1}(A)}(x,y)
$$
From what you mentioned in the question we also know that $\mu(\gamma^{-1}A)=\frac{1}{|a||b|}\mu(A)$.
So
$$\int_\mathbb {R^2}r(\gamma(x,y))d\mu=\int_\mathbb {R^2}1_A(\gamma(x,y))d\mu=\int_\mathbb {R^2}1_{\gamma^{-1}(A)}(x,y)d\mu=\mu(\gamma^{-1}A)$$$$=\frac{1}{|a||b|}\mu(A)=\frac{1}{|a||b|}\int_{\mathbb{R^2}}r(x,y)d\mu$$

Hence we have the inequality here
  $$\int_\mathbb{R^2} \phi(x,y)d\mu \le \int_{\mathbb{R^2}}r(x,y)d\mu=|a||b|\int_\mathbb {R^2}r(\gamma(x,y))d\mu
$$


So we have proved our claim for characteristic functions .
Then it is also true for simple functions $r$. Observe the key is to get the equality

$$ \int_{\mathbb{R^2}}r(x,y)d\mu=|a||b|\int_\mathbb {R^2}r(\gamma(x,y))d\mu$$

You have this for simple function now.
We know that any positive measurable function can be written as increasing limit of simple functions (point-wise limit). Then by monotone convergence you have the equality. Hence you are done.
