The domain of the Haar measure of the $n$-dimensional Torus

Question

I'm trying to find the Haar measure of the $n$-dimensional Torus. To find this measure I'm using the notion of pushforward measure of the Lebesgue measure.

To understand my question please consider the information below:

Let $\mathbb{T}^n:=\mathbb{R}^n/\mathbb{Z}^n$ be the quotient of the group $(\mathbb{R}^n,+)$ by the subgroup $(\mathbb{Z}^n,+)$. Define $d _{\mathbb{T}^n}:\mathbb{T}^n\times \mathbb{T}^n\to\mathbb{R}$ by $d_{\mathbb{T}^n}(x+\mathbb{Z}^n,y+\mathbb{Z}^n) :=\inf \big\{\Vert x-y+ k\Vert _n:k\in\mathbb{Z}^n\big\}$.

We can show that $(\mathbb{T}^n,d _{\mathbb{T}^n})$ is a metric space and that $\mathbb{T}^n$ is a abelian compact group with respect to that metric. We can also show that $\pi :\mathbb{R}^n\to \mathbb{T}^n$ given by $\pi (x):=x+\mathbb{Z}^n$ is an open Lipschitz continuous map.

Define $\pi_\star \mathfrak{B}_n :=\big\{B\subseteq\mathbb{T}^n:\pi ^{-1}[B]\in \mathfrak{B}_n\big\} $ in which $\mathfrak{B}_n$ is the Borel $\sigma$-algebra of $\mathbb{R}^n$.

Suppose that $\mathfrak{B}_{\mathbb{T}^n}$ is the Borel $\sigma$-algebra of $(\mathbb{T}^n,d _{\mathbb{T}^n})$.

My question is: does the equality $\mathfrak{B}_{\mathbb{T}^n}=\pi _\star \mathfrak{B}_n$ hold?

It's easy to show that $\mathfrak{B}_{\mathbb{T}^n}\subseteq \pi _\star \mathfrak{B}_n$ since $\pi$ is a measurable map. However I don't know if $\pi_\star \mathfrak{B}_n\subseteq \mathfrak{B}_{\mathbb{T}^n}$.

Thank you for your attention.

Answer 1

The answer is yes. For the natural projection $\pi:\mathbb{R}^n\to\mathbb{T}^n$, the measurable subset $\Sigma=[0,1[^n\subseteq \mathbb{R}^n$ is a strict fundamental domain (i.e. $\pi|_\Sigma:\Sigma\to \mathbb{T}^n$ is a bimeasurable bijection).

Let $B\subseteq \mathbb{T}^n$ be a subset such that $\pi^{-1}(B)\in\mathfrak{B}({\mathbb{R}^n})$. Then

$$B=\pi|_\Sigma((\pi|_\Sigma)^{-1}(B))= \pi|_\Sigma(\Sigma\cap \pi^{-1}(B)).$$

Since $\Sigma$ and $\pi$ are measurable, so is $\Sigma\cap \pi^{-1}(B)$. Since $(\pi|_\Sigma)^{-1}$ is measurable so is $B$. Thus $\pi_\ast(\mathfrak{B}(\mathbb{R}^n))\leq \mathfrak{B}(\mathbb{T}^n)$.

See https://mathoverflow.net/q/50541/66883 for a more general version of this argument. Also relevant is Characterization of the Borel $\sigma$-algebra on a topological quotient space.