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 $\mathcal{B}(\mathbb{T})$ ?
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 09 '12 at 19:07(1) I showed that $I:={ M \in \mathcal{B}(\mathbb{T}) ; \mu(aM) = \mu(M) \forall a \in \mathbb{T}}$ is a Dynkin-System.
(2) I showed (as Chris proposed) that the open balls from the subspace topology on $\mathbb{T}$ are elements of $I$. Moreover they are a generator of $ \mathcal{B}(\mathbb{T}) $ which is $\cap$-stable. (3) Thus we have $I = \mathcal{B}(\mathbb{T}) $. I hope this is right now. Thanks for your help!
– Oscar Mar 09 '12 at 23:03