Prove that $\lambda(\alpha A) = \alpha\lambda(A)$ where $\lambda$ is the Lebesgue measure and $A$ is a Borel-Set

Question

Let $\mathscr{B}(\mathbb{R})$ be the Borel-Set over $\mathbb{R}$ and $\lambda: \mathscr{B}(\mathbb{R}) \to [0,\infty]$ the Lebesgue measure.

Prove: If $A \in \mathscr{B}(\mathbb{R})$, show that $\alpha A := \{\alpha x \; |\; x \in A\}\in \mathscr{B}(\mathbb{R})$ for $\alpha > 0$ and further show that $\lambda(\alpha A) = \alpha \lambda(A)$.

I know that by definition, $\mathscr{B}(\mathbb{R})$ is the smallest $\sigma$-Algebra that contains all open sets in $\mathbb{R}$. I'm confused how to derive $\alpha A \in \mathscr{B}(\mathbb{R})$ from $A \in \mathscr{B}(\mathbb{R})$. Any hints are greatly appreciated.

Answer 1

It is clear that if $I\subset\Bbb R$ is an interval, then $\alpha I$ is an interval for every $\alpha>0$, and that ${\rm long\,}(\alpha I)=\alpha {\rm long\,}(I)$.

If $A\subset \bigcup I_n$ with $I_n$ intervals, then $\alpha A\subset \bigcup (\alpha I_n)$ and $\alpha I_n$ are intervals. So, $$\lambda(\alpha A)\le \sum_n{\rm long\,}(\alpha I_n)=\alpha \sum_n{\rm long\,}(I_n).$$ Now, taking $\inf$ over all covers of $A$, and taking into account that $\alpha>0$, you get that $$\lambda(\alpha A)\le \alpha \lambda (A).$$

For the reverse, $\alpha \lambda(A)=\alpha \lambda[(1/\alpha) (\alpha A)]\le\alpha (1/\alpha) \lambda(\alpha A)=\lambda(\alpha A)$.