On translation and dilation invariant Lebesgue measure: Folland's text

[psie]

TL;DR

I am stuck at a claim made in Theorem 1.21 in Folland's real analysis text on the translation and invariant Lebesgue measure on $\mathbb R$.

Let $m$ be Lebesgue measure and $\mathcal L$ the Lebesgue $\sigma$-algebra (the complete $\sigma$-algebra that includes the Borel $\sigma$-algebra). Consider,

Theorem 1.21. If $E\in\mathcal L$, then $E+s\in\mathcal L$ and $rE\in \mathcal L$ for all $s,r\in \mathbb R$. Moreover, $m(E+s)=m(E)$ and $m(rE)=|r|m(E)$.

Folland starts off by saying that the collection of open intervals is invariant under translations and dilations, so the same is true for $\mathcal B_\mathbb{R}$. I understand that claim; basically consider $\mathcal B'=\{E\in\mathcal B_\mathbb{R}:E+t\in\mathcal B_\mathbb{R},t\in\mathbb R\}$ and show it's a $\sigma$-algebra that contains all the open intervals and hence $\mathcal B_\mathbb{R}$, and by definition, $\mathcal B'\subset \mathcal B_\mathbb{R}$. So $\mathcal B'=\mathcal B_\mathbb{R}$. Same for dilations.

However, then he goes on to say that if we let $m_s(E)=m(E+s)$ and $m^r(E)=m(rE)$ for $E\in \mathcal B_\mathbb{R}$, then they "clearly" agree with $m$ and $|r|m$ on finite unions of intervals. This claim I don't understand. If we define $\phi_t(x)=x+t$ as the translation by $t$, and if $A$ is a finite union of intervals, then $$\phi_t(A)=\phi_t\left(\bigcup_{n=1}^N E_n\right) = \bigcup_{n=1}^N \phi_t(E_n).$$ But how do I show $m_s(A)=m(A)$? The thing that's confusing me is that up until now, he has only talked about finite disjoint union of h-intervals, where an h-interval is a set of the form $(a,b]$, $(a,\infty)$ or $\varnothing$ for $-\infty\leq a<b<\infty$. I'm sure his statement is correct as it stands, but I don't know how to show $m_s(A)=m(A)$.

 

[pasmith]

What's the measure of (a,b] + s = (a + s, b+ s]?
What's the measure of r(a,b] = \begin{cases}<br /> (ra, rb] &amp; r &gt; 0 \\<br /> (0,0] &amp; r = 0 \\<br /> [rb, ra) &amp; r &lt; 0 \end{cases}

 

[psie]

Thanks @pasmith! I'll just add this here as an answer to my question.

The author notes prior to the theorem that the Lebesgue measure of an interval is just its length, so if $I$ is an interval with endpoints $a,b$ with $a<b$, its measure is $b-a$. Clearly $m_s(I)=m(I)$ and $m^r(I)=|r|m(I)$. And if $A=\bigcup_1^n I_i$ where $I_i$ are intervals, then first observe that we can make them disjoint and obtain $A=\bigcup_1^k J_j$. Disjoint is preserved under translation and dilation, so if $\phi_s(x)=x+s$ (the very same argument applies to $\phi_r(x)=rx$), it follows that \begin{align*}m_s(A)&=m(\phi_s(A))=m\left(\phi_s\left(\bigcup_{j=1}^k J_j\right)\right)= m\left(\bigcup_{j=1}^k \phi_s(J_j)\right)=\sum_1^k m(\phi_s(J_j))\\ &=\sum_1^k m_s(J_j)=\sum_1^k m(J_j)=m\left(\bigcup_{j=1}^kJ_j\right)=m(A).\end{align*} Thus $m_s(E)=m(E)$ and $m^r(E)=|r|m(E)$ for $E$ being a finite union of intervals.