$ \varphi(x_1, \dots , x_n) = (c_1x_1, \dots, c_nx_n)$. Then, $\mu(\varphi(A)) = |c_1\dots c_n|\mu(A)$, where $\mu$ is the Lebesgue measure

Question

Given $c_1, \dots, c_n \in \mathbb{R}$, consider the function $\varphi: \mathbb{R}^n \rightarrow \mathbb{R}^n, \varphi(x_1, \dots , x_n) = (c_1x_1, \dots, c_nx_n)$. If $\mu$ is the Lebesgue measure, show that for every $A \in \mathcal{M}(\mathbb{R})$ (set of Lebesgue measurable sets), we have \begin{equation*} \mu(\varphi(A)) = |c_1\dots c_n|\mu(A) \end{equation*}

First, notice that $\varphi(A) \in \mathcal{M}(A)$.

\begin{align*} \mu(\varphi(A)) &= \int_{\varphi(A)} 1 dm(x)\\ &= \int_{\mathbb{R}^n} \chi_{\varphi(A)} dm(x) \end{align*}

Now, notice that $x = (x_1, \dots, x_n) \in \varphi(A)$ iff $x_i = c_iy_i$, for some $y = (y_1, \dots, y_n)$.

I don't know how to put this information inside the integral so that I can conclude the result. Could someone help me?

Answer 1

Fill in the details in the following argument.

If $c_i = 0$ for some $i$, then $\varphi(A)$ is contained in a hyperplane of $\Bbb R^n$, which has measure zero. Then $\varphi(A)$ has measure zero and the result is clear. Now suppose none of the $c_i$ are zero. Fix a positive number $\varepsilon$, and let $\{R_j\}$ be a covering of $A$ by open rectangles such that $$\sum \operatorname{vol}(R_j) < \mu(A) + \frac{\varepsilon}{\lvert c_1\cdots c_n\rvert}$$ Note that $\{\varphi(R_j)\}$ is a covering of $\varphi(A)$ by open rectangles, and for every $j$, $$\operatorname{vol}(\varphi(R_j)) = \lvert c_1\cdots c_n\rvert \operatorname{vol}(R_j)$$ Furthermore, $$\mu(\varphi(A)) \le \sum \operatorname{vol}(R_j) =\lvert c_1\cdots c_n\rvert \sum \operatorname{vol}(R_j) < \lvert c_1\cdots c_n\rvert\mu(A) + \varepsilon$$ On the other hand, let $\{G_j\}$ be a covering of $\varphi(A)$ by open rectangles such that $\sum \operatorname{vol}(R_j) < \mu(\varphi(A)) + \epsilon$. Then $\{\varphi^{-1}(G_j) \}$ is a covering of $A$ by open rectangles with $\lvert c_1\cdots c_n\rvert \operatorname{vol}(\varphi^{-1}(G_j)) = \operatorname{vol}(G_j)$ for all $j$, and $$|c_1\cdots c_n|\mu(A) \le \sum\lvert c_1\cdots c_n\rvert \operatorname{vol}(\varphi^{-1}(G_j)) = \sum \operatorname{vol}(G_j) < \mu(\varphi(A)) + \varepsilon$$ Thus $$\mu(\varphi(A)) - \varepsilon < \lvert c_1\cdots c_n\rvert \mu(A) < \mu(\varphi(A)) + \varepsilon$$ Since $\epsilon$ was arbitrary, the result follows.