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Good evening!

Let $ \mathbb{T}:=\{ z \in \mathbb{C} ; \vert z \vert =1 \} $ be the unit circle in the complex plane. We denote the trace Borel-$\sigma$-algebra on $\mathbb{T}$ by $\mathcal{B}(\mathbb{T})$.

Here is my question: Does anyone know an elegant method to construct the Haar-measure (in this case the one-dimensional Lebesgue-measure) on $\mathcal{B}(\mathbb{T})$ ?

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It seems to me that almost any construction of Lebesgue measure on $\mathbb{R}$ or $[0,1]$ would work on the circle with trivial modifications. Is there something special you're looking for? –  Nate Eldredge Mar 9 '12 at 19:03
    
Hi Nate! First I used this construction: First I defined $ f: [0,1) \to \mathbb{T}~ ; ~ t \mapsto \exp(2\pi i t)$. The Haar measure is now given by: $\mu = \lambda \circ f^{-1}$. With this construction it is not so nice to show that $\mu$ is invariant under the group operation... thats why i was looking for a better construction. with best regards, Oscar –  Oscar Mar 9 '12 at 19:07
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The construction you gave there is the usual one. What is the trouble with sets modulo [0,1)? By the monotone class theorem (or Pi-Lambda) it suffices to consider the image of intervals (or balls in the subspace topology on the torus) where showing invariance is not too hard. –  Chris Janjigian Mar 9 '12 at 19:37
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If you have a Borel set on the unit circle, you simply "cut" the circle and measure the Borel set in reals... –  N. S. Mar 9 '12 at 20:00
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To elaborate on N.S.'s comment, your $\mu = \lambda \circ f^{-1}$ construction would work if you could show that Lebesgue measure $\lambda$ on $[0,1)$ is invariant under translation mod 1. But this is easy if you just note that $A + t \bmod 1 = ((A \cap [0,1-t)) + t) \cup ((A \cap [1-t,1)) - (1-t))$ where the union is disjoint. –  Nate Eldredge Mar 9 '12 at 23:58

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