Constructing the Haar measure of the $n$-dimensional Torus

borel-measuresborel-setshaar-measurelebesgue-measuremeasure-theory

Let $\mathbb{T}^n:=\mathbb{R}^n/\mathbb{Z}^n$ be the quotient of the group $(\mathbb{R}^n,+)$ by the subgroup $(\mathbb{Z}^n,+)$.

I'm trying to construct the Haar measure of $\mathbb{T}^n$. I constructed a measure which is finite and Radon. However I don't how to prove that the measure I constructed is translation invariant. Below I'll sketch the construction. Firstly suppose that $\lambda$ is the Lebesgue measure of $\mathbb{R}^n$ and that $\mathfrak{B}$ is the Borel $\sigma$-algebra of the unit cube $[0,1]^n$ with the subspace topology of $\mathbb{R}^n$.

  • Define $\pi :[0,1]^n\to \mathbb{T}^n$ by $\pi (x):=x+\mathbb{Z}^n$. We can define a metric $d_{\mathbb{T}^n}$ in $\mathbb{T}^n$ such that $\pi$ is continuous (see here).
  • Let $\mathfrak{B}_{\mathbb{T}^n}$ be the Borel measure of $\mathbb{T}^n$ induced by the topology induced by the metric $d_{\mathbb{T}^n}$;
  • Let $\pi _\star (\lambda |_{\mathfrak{B}}) :\pi_\star\mathfrak{B}\to\overline {\mathbb{R}} $ be the pushforward measure of the restriction $\lambda|_{\mathfrak{B}} $. It's easy to show that $\mathfrak{B}_{\mathbb{T}^n}\subseteq\pi_\star\mathfrak{B} $ since $\pi$ is continuous.

Define $\vartheta :\mathfrak{B}_{\mathbb{T}^n}\to \overline{\mathbb{R}}$ by $\vartheta (B):=\pi _\star (\lambda |_{\mathfrak{B}}) (B)$.

We can show that $\vartheta :\mathfrak{B}_{\mathbb{T}^n}\to \overline{\mathbb{R}}$ is a probability measure of $\mathbb{T}^n$ and it's also a Radon measure.

My question is: how can I show that $\vartheta $ is translation invariant?

I did almost nothing worth mentioning.

I think it's important to observe that the restriction $\pi |_{[0,1)^n}:[0,1)^n\to\mathbb{T}^n$ is bijective.

With the property of pushforward measure we can show that, given any $B\in\mathfrak{B}_{\mathbb{T}^n}$ and $a\in\mathbb{T}^n$, we have

$$\vartheta(B+a)=\int _{\mathbb{T}^n}\mathbf{1}_{B+a}d\vartheta=\int _{[0,1]^n}\mathbf{1}_{B+a}(\pi (x))d\lambda (x)\,\,\color{red}{(1)} $$

Above $\mathbf{1}_{B+a}$ is the indicator function of $B+a$.

If there's an $\alpha\in[0,1]^n$ such that $\mathbf{1}_{B+a}(\pi (x))=\mathbf{1}_{\pi^{-1}(B)+\alpha}(x)$ for all $x\in [0,1]^n$, then we can use $(1)$ to prove that $\vartheta (B+a)=\lambda (\pi ^{-1}(B)+\alpha )=\lambda (\pi^{-1}(B))=\vartheta(B)$. However I wasn't able to prove that there's such $\alpha\in[0,1]^n$.

If anyone knows an easier way to construct the Haar measure of the $n$-dimensional Torus please tell me.

Thank you for your attention!

Best Answer

One slightly indirect way to do this is to invoke the Riesz-Markov-Kakutani theorem, with corollary that, given a discrete subgroup $\Gamma$ (thinking of $\mathbb Z$) of an abelian topological group $G$ (thinking of $\mathbb R$), fixing a Haar measure $\mu$ on $G$, there is a unique measure $\nu$ on $G/\Gamma$ such that $$ \int_{G/\Gamma} \sum_{\gamma\in\Gamma} f(\gamma+\dot{g})\;d\nu(\dot{g}) \;=\; \int_G f(g)\;d\mu(g) $$ As part of this, one proves that the averaging map $C^o_c(G)\to C^o(G/\Gamma)$ is surjective.

This characterization can be used to prove essentially all of the desired properties, such as group-invariance on the quotient $G/\Gamma$, from the group-invariance on $G$.

EDIT: the proof of the corollary is short and straightforward: given $f\in C^o(G/\Gamma)$, let $F\in C^0_c(G)$ be such that $f(g) = \sum_{\gamma\in G/\Gamma} F(\gamma+g)$. For fixed $\dot{g}_o\in G/\Gamma$, let $g_o\in G$ be a representative. Then the averaged version of $F(g+g_o)$ is $f(\dot{g}+\dot{g_o})$, and $$ \int_{G/\Gamma}f(\dot{g}+\dot{g}_o)\,d\dot{g} \;=\; \int_G F(g+g_o)\;dg \;=\; \int_G F(g)\;dg \;=\; \int_{G/\Gamma} f(\dot{g})\;d\dot{g} $$ Here we use the important fact that the corresponding integral is translation-invariant if and only if the measure is translation-invariant. :)

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