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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.

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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 $U\subset \mathbb R.$

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

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  • $\begingroup$ Hi thanks for the comment, I was thinking about this approach at first but had some concerns about the fact that it's not neccesary that $a_n$ and $b_n$ are rationals, since only their difference need be rational - is there still a way to make this method work? $\endgroup$ – UsernameInvalid Nov 21 '18 at 22:09
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    $\begingroup$ But if both $a_n,b_n$ are rational, you have it covered. $\endgroup$ – zhw. Nov 21 '18 at 22:11
  • $\begingroup$ Ah yes of course, I suppose I was looking into it too much, thanks for the help :) $\endgroup$ – UsernameInvalid Nov 21 '18 at 22:14

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