Lebesgue measure is translation invariant and scales by $\delta^n$ under dilation.Suggest a slogan
Let \(m\) be Lebesgue measure on \(\RR^n\), let \(E\subseteq\RR^n\) be Lebesgue measurable, let \(h\in\RR^n\), and let \(\delta>0\). Then \(E+h\coloneqq\theset{x+h\suchthat x\in E}\) and \(\delta E\coloneqq\theset{\delta x\suchthat x\in E}\) are Lebesgue measurable, and \[ m(E+h)=m(E), \qquad m(\delta E)=\delta^n\,m(E). \]