Understanding the proof $\mu$ is invariant then $\mu$ is a linear transformation of Lebesgue measure

Question

Exercise: Let $\mu$ be a Lebesgue-Stieltjes measure on $\mathscr{B}_{\mathbb{R}}$ invariant for the class of right half-closed intervals of $\mathbb{R}$, so that, $\mu(a+I)=\mu(I)$, for all $a\in\mathbb{R}$ and $I=(x,y]$. Show that, in $\mathscr{B}_\mathbb{R}$, $\mu=c.Leb$ where c\in$\mathbb{R}$ and Leb denotes the Lebesgue measure.

I posted this question on another thread and this answer from another thread was suggested. Due to the fact it is an old post I did not expect the author to answer me:

The answer was:

"Here is a way to argue out. I will let you fill in the details.

1. If we let $\mu([0,1))=C$, then $\mu([0,1/n)) = C/n$, where $n \in \mathbb{Z}^+$. This follows from additivity and translation invariance.
   2. Now prove that if $(b-a) \in \mathbb{Q}^+$, then $\mu([a,b)) = C(b-a)$ using translation invariance and what you obtained from the previous result.
   3. Now use the monotonicity of the measure to get lower continuity of the measure for all intervals $[a,b)$.

Hence, $\mu([a,b)) = \mu([0,1]) \times(b-a)$." by user17762

Attempted proof: 1) It is true the $[0,1]=\bigcup_{i=0}^{n}(\frac{i}{n},\frac{i+1}{n}]$

Since the measure $\mu$ is invariant then $\mu((\frac{i}{n},\frac{i+1}{n}])=\mu((\frac{i}{n}-\frac{1}{n},\frac{i+1}{n}-\frac{1}{n}])=\mu((\frac{i-1}{n},\frac{i}{n}])$, which proves every individual set of the covering has the same measure then by addititvity $\mu((0,1])=\mu(\bigcup_\limits{i=0}^{n}(\frac{i}{n},\frac{i+1}{n}])=\sum_\limits{i=0}^{n}\mu((\frac{i}{n},\frac{i+1}{n}])=\mu((0,1])$

This implies $\mu((0,\frac{1}{n}])=\frac{C}{n}$. However I am having trouble on proving 2) once I cannot relate the interval (a,b] and its respective length to the previous definition as the author intended.

Question:

Can someone help me prove point 2) and explain point 3)?

Thanks in advance!

Answer 1

Although you have another kind of intervals in your main question, you can easily "reverse" your proof to get $\mu((0,1/n])=C/n$ with $C=\mu((0,1])$. You are almost there, let me help you with part 2 first. For the measure of $(a, b] $ with $a, b\in\mathbb Q$ it is enough to consider the case with nonnegative rationals. Indeed we can always consider $(a, b] - a$ otherwise. One has by the exclusion property of the measure \begin{align}\tag{$*$}\mu((a,b])=\mu((0,b]\setminus(0,a])=\mu((0,b])-\mu((0,a])\end{align} So it is enough to show that for $a=p/q$ with $p,q$ nonnegative integers \begin{align} \mu((0,a])=Cp/q \end{align} We write \begin{align} \mu((0,a])=\mu\left(\bigcup_{k=1}^p \left(\frac{k-1}{q},\frac{k}{q}\right] \right) = \sum_{k=1}^p \mu\left(\left(\frac{k-1}{q},\frac{k}{q}\right] \right) = \sum_{k=1}^p \mu\left(\left(0,\frac{1}{q}\right] \right)=Cp/q=Ca \end{align} Since $a$ was arbitrary choice the same holds for $b\in\mathbb Q$ implying that equation $(*)$ can be written as \begin{align} \mu((a,b])=\mu((0,b])-\mu((0,a])=Cb-Ca=C(b-a) \end{align} I finish the proof in a (slightly) different way. Notice that \begin{align} \mathcal C:=\{(a,b]\ :\ a,b\in \mathbb Q\} \end{align} is a $\pi$-system that generates the Borel $\sigma$-algebra. We have just showed that \begin{align} C\operatorname{Leb}(A)=\mu(A) \end{align} for all $A\in\mathcal C$. By the uniqueness of measure, we conclude that \begin{align} C\operatorname{Leb}(A)=\mu(A) \end{align} for all $A\in\mathcal B$.

Answer 2

I think the proof will be easier to push through if we break it up a bit. If this answer is not what you are looking for, I will be glad to delete it. But all you really need to know here is that every open set is a countable disjoint union of half-open intervals with rational endpoints (even dyadic rational endpoints). Then, the main idea is that Lebesgue measure is the only translation-invariant measure on $\mathscr B(\mathbb R)$ that assigns to each half-open interval with rational endpoints, its length. From there, it's a little trick to finish the proof.

Cohn does it like this:

Suppose that $\mu$ is another measure that does so. Then, if $U$ is an open subset of $\mathbb R$, it is a disjoint countable union of half-open intervals with rational endpoints $I_n$. Then,

$\mu (U)=\sum \mu (I_n)=\sum \lambda (I_n)=\lambda (U).$

So, $\mu$ and $\lambda$ agree on the open sets. Regularity of $\lambda$ now implies that $\mu(E)\le \lambda(E)$ for all Borel sets.

For the reverse inequality, suppose that $A$ is a bounded Borel set and take an open set $V$ containing $A$ and apply the previous inequality, to get

$\mu(V)=\mu (A)+\mu (V-A)\le \lambda (A)+\lambda (V-A)=\lambda (V)$

so $\mu(A)=\lambda (A).$

For the unbounded case, note that $A=\cup_n (-n,n]\cap A$ and use the countable additivity of $\mu$ and $\lambda$.

So, $\mu=\lambda$ on the Borel sets, which proves the main claim.

To finish, define a new measure $\nu$ on the Borel sets of $\mathbb R$ by $\nu(E)=\frac{1}{c}\mu(E)$. Then, $\nu$ is translation invariant, and $\nu((0,1])=\lambda((0,1]).$

Take an interval $I=(r,r+2^{-k}];\ r\in \mathbb Q.$ Then, $I$ is an interval with rational endpoints, and now, using the translation invariance the measures, and the result we just proved, we have

$2^k\cdot \nu(I)=\nu((0,1])=\lambda((0,1])=2^k\lambda (I)\Rightarrow \nu(I)=\lambda(I)$,

so $\nu=\lambda$ on the Borel sets, and thus $\mu=c\lambda.$