# Show that a given measure is equal to the Lebesgue measure on Borel subsets on $\mathbb{R}$

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 $$a Proof: Write $$(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.$$
• 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? – UsernameInvalid Nov 21 '18 at 22:09
• But if both $a_n,b_n$ are rational, you have it covered. – zhw. Nov 21 '18 at 22:11