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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 $ainmathbb{R}$ and $I=(x,y]$. Show that, in $mathscr{B}_mathbb{R}$, $mu=c.Leb$ where cin$mathbb{R}$ 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!

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 $a, binmathbb Q$ 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])=muleft(bigcup_{k=1}^p left(frac{k-1}{q},frac{k}{q}right] right) = sum_{k=1}^p muleft(left(frac{k-1}{q},frac{k}{q}right] right) = sum_{k=1}^p muleft(left(0,frac{1}{q}right] right)=Cp/q=Ca
end{align}
Since $a$ was arbitrary choice the same holds for $binmathbb Q$ 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,bin mathbb Q}
end{align}
is a $pi$-system that generates the Borel $sigma$-algebra. We have just showed that
begin{align}
Coperatorname{Leb}(A)=mu(A)
end{align}
for all $Ainmathcal C$. By the uniqueness of measure, we conclude that
begin{align}
Coperatorname{Leb}(A)=mu(A)
end{align}
for all $Ainmathcal B$.

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

$2^kcdot nu(I)=nu((0,1])=lambda((0,1])=2^klambda (I)Rightarrow nu(I)=lambda(I)$,

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