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

lebesgue-measuremeasure-theoryreal-analysis

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.
    1. 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.
    2. 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!

Best Answer

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 $A\in\mathcal B$.

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