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Suppose we have a measure $mu$ on $left( mathbb{R},mathcal{B}(mathbb{R}) right)$ with $mu((0,1])=1$, and $mu$ invariant under translations i.e. $mu(A) = mu(A+c)$ for every $c in mathbb{R}$ and Borel set $A subseteq mathbb{R}$.

Now letting $lambda$ be the Lebesgue measure, in the first part of the question I was able to show that for any interval $A=(a,b]$ with $b-a in mathbb{Q}$ that $mu(A) = lambda(A)$.

The second part of the question asks to extend this to all Borel sets, i.e. $forall A in mathcal{B}(mathbb{R})$ we get that$mu(A) = lambda(A)$, but I am really struggling to think of a practical way of achieving this.

Clearly using the first part of the question is the correct approach, so I was attempting to come up with some ways of using this.

The first thing I thought of was maybe to approximate all of the intervals in $mathcal{B}(mathbb{R})$ by these rational intervals in the first part of the question, but I could not think of a rigorous way of doing this and I do not think this is the correct approach.

My next thought was to potentially show that the intervals from the first part form a $pi$-system of some sort and maybe generate the Borel sets and so agreeing only on these intervals is sufficient to show they agree on the whole space, but I am not too confident in arguing this so any help would be appreciated thanks.

Having $mu (a,b]=b-a$ for rational $a,b$ gives $mu (a,b)=b-a$ for all real $a<b.$ Proof: Write $(a,b)$ as the increasing union of $(a_n,b_n]$ for appropritate rational $a_n,b_n.$ Standard measure theory with your result for rationals then gives the result.

Since every open set in $mathbb R $ is the disjoint union of open intervals, we see $mu(U) = lambda (U)$ for all open $Usubset mathbb R.$

I'll stop here for now. Ask questions if you like.