Spectral theorem for representations proof. Let $H$ be a separable Hilbert space, and $U$ a unitary representation of $\mathbb{Z}^d$ on $H$. Let $\chi_m$ be the characters of the Torus $T^d$, and $m$ the Haar measure on $T^d$. I would like to Show that if $U$ is cyclic with cyclic vector $v$, then there is a finite measure $\mu_v$ on $T^d$ such that $\langle U(m)v,v\rangle=\int_{T^d}\chi_m d\mu_v$. I would like to do this by considering the sequence of measure on $T^d$ defined by $d\mu_{v,N}=\frac{1}{(N+1)^d}||\sum_{m\in\mathbb{Z}^d, m_i\leqslant N}\chi_m^{-1} U(m)v||^2 dm$. 
I guess we should pass to the Limit $T\rightarrow\infty$, but I am not too sure how to do it in Detail.
Thanks for your help.

(note that you are using $m$ for two different things; I will just ignore the Haar measure as I don't think it is needed)
Since the characters $\chi_m$ evaluate at single points $m\in\mathbb Z^d\cap\mathbb T^d$ (which are the $d$-tuples consisting of $1$ and $-1$) , you have
$$
\int_{\mathbb T^d}\chi_m\,d\mu_v=\mu_v(\{m\}).
$$
And you want this to equal $\langle U(m)v,v\rangle$. So it is clear what the measure $\mu_v$ should be:
$$
\mu_v(X)=\sum_{m\in X\cap\mathbb Z^d}\,\langle U(m)v,v\rangle\,\delta_m.
$$
Then, for a fixed $m\in\mathbb Z^d\cap\mathbb T^d$,
$$
\int_{\mathbb T^d}\chi_m\,d\mu_v=\mu_v(\{m\})=\langle U(m)v,v\rangle.
$$
