# Lebesgue measure on $\mathbb{R}/\mathbb{Z}$

I was reading a (brief) introduction about measure theory today and came across the following statement:

(Lebesgue measure on $\mathbb{R}/\mathbb{Z}$): There is a unique probability measure $\mu$ on $\mathbb{R}/\mathbb{Z}$ such that $\mu((a,b))=b-a$ for all $0\le a<b\le 1$.

My question is not about the proof (the notes are too brief to have the proofs) but about the $\sigma$-algebra in question. I assume that $(\mathbb{R}/\mathbb{Z},B,\mu)$ is a measure space where $B$ is the Borel $\sigma$-algebra, i.e. the smallest $\sigma$-algebra containing all the open balls. What is the metric in question? How come all elements of $B$ of the form $(a,b)$? Do we use the bijection between $\mathbb{R}/\mathbb{Z}$ and $[0,1)$ somewhere?

I'll be grateful if someone can clarify my doubts. Thanks.

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It is not true that every element of $B$ is an open interval. Rather, the open intervals generate $B$ as a $\sigma$-algebra. – Zhen Lin Jul 10 '12 at 11:10
Why is any element of $B$ an open interval? Aren't elements of $B$ subsets of $\{x+\mathbb{Z}:x\in\mathbb{R}\}$? – Shahab Jul 10 '12 at 11:15
@Shahab: There's a canonical injection $[0,1)\to \mathbb R/\mathbb Z$. The image of an open interval under this bijection is an element of $B$. Alternatively, you can consider $\mathbb R/\mathbb Z$ to be $[0,1)$ -- albeit with a nonstandard topology -- simply by always choosing the unique representative for each class that lies in this interval. – Henning Makholm Jul 10 '12 at 11:25

Yes, one uses the bijection you mention. To be precise, let $f:[0,1]\to S^1$ be the defining quotient map $t\mapsto e^{2\pi i t}$. Write $\lambda$ for Lebesgue measure on $[0,1]$.
Then the property is $\mu(S)=\lambda(f^{-1}(S))$, or $\mu(f(a,b))=\lambda(a,b):=b-a$. This is usually called Haar measure (rather than Lebesgue measure) on $S^1$.