Let $f$ be a real-valued integrable function on $\mathbb{R}^d$. Prove that $$\int f(\delta x) = \delta^{-d} \int f.$$
I let $f(x)=\chi_E(x)=\begin{cases} 1 & \text{if }\delta x \in E \\ 0 & \text{if }\delta x \in E \end{cases} = \begin{cases} 1 & \text{if } x \in \delta^{-1} E \\ 0 & \text{if } x \in \delta^{-1}E \end{cases}$.
Putting this together, I start with \begin{align*} m(\delta^{-1}E)&=\delta^{-d}m(E) \\ \sum_{k=1}^N a_k m(\delta^{-1}E)&=\delta^{-d} \sum_{k=1}^Na_km(E) \\ \end{align*} And as $N \to \infty$, \begin{align*} \int \chi_{\delta^{-1}E}(x)&= \delta^{-d} \int \chi_E(x) \\ \int f(\delta x) &= \delta^{-d} \int f(x). \end{align*} Is this correct?