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

  
*
  
*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. 
  
  
  
*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.
  
*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! 
 A: 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 $\mu=c\lambda.$
A: 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 $A\in\mathcal B$. 
