Show $\int_{\mathbb R^{d}}f(\alpha x)d\lambda^{d}(x)=\frac{1}{|\alpha|^{d}}\int_{\mathbb R^{d}}f(x)d\lambda^{d}(x)$

Question

Let $\alpha \neq 0$

Use $|\alpha|^{d}\lambda^{d}(B)=\lambda^{d}(\alpha B)$ $(*)$ to:

Show that $\int_{\mathbb R^{d}}f(\alpha x)d\lambda^{d}(x)=\frac{1}{|\alpha|^{d}}\int_{\mathbb R^{d}}f(x)d\lambda^{d}(x)$

Ideas:

Let $f\geq0$ and measurable then we can find $(g_{n})_{n}$ of step functions that approximate $f$, i.e. $\sup g_{n}:=g=f$. Therefore

$\int_{\mathbb R^{d}}f(\alpha x)d\lambda^{d}(x)=\int_{\mathbb R^{d}}g(\alpha x)d\lambda^{d}(x)$ and by monotone convergence, we get:

$\int_{\mathbb R^{d}}g(\alpha x)d\lambda^{d}(x)=\sup_{g_{n} \in E, 0 \leq g_{n} \leq f}\int_{\mathbb R^{d}}\sum_{i=1}^{N}\alpha_{i}\chi_{A_{i}}(\alpha x)d\lambda^{d}(x)=\sup_{g_{n} \in E, 0 \leq g_{n} \leq f}\sum_{i=1}^{N}\alpha_{i}\lambda^{d}(A_{i})$

and using $(*)$ this means $\sum_{i=1}^{N}\alpha_{i}\lambda^{d}(A_{i})=|\alpha|^{d}\sup_{g_{n} \in E, 0 \leq g_{n} \leq f}\sum_{i=1}^{N}\alpha_{i}\lambda^{d}(\alpha^{-1}A_{i})$

But this leads nowhere...

Any ideas?

Answer 1

You made a small mistake when integrating the indicator function. $$\int \chi_{A_i}(\alpha x) \, d\lambda^d(x) = \lambda^d(A_i/\alpha) = |\alpha|^{-d} \lambda^d(A_i).$$