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.
 A: Let's fix some notations.
Denote the Haar measure on $\Bbb{T}^d$ by $\lambda$ instead of $m$.
The measures suggested should probably be normalized by
$$ d\mu_{v,N}(y)=\frac{1}{(2N+1)^d}\left\|\sum_{m\in\mathbb{Z}^d, \lvert
m_i \rvert \leq N}\chi_{-m}(y) U(m)v\right\|^2 d\lambda(y).$$
The characters of $\Bbb{T}^d$ are the functions
$$\chi_m: \Bbb{T}^d  \to C^*\\
y \mapsto e^{2\pi i m \cdot y}$$
Finally, introduce the shorthand notation for the truncated sums
$$\sum_{m \le N} f(m) := \sum_{m\in\mathbb{Z}^d, \lvert m_i\rvert \leq N} f(m).$$
With this, we have
\begin{align*}
\left\| \sum_{m\le N} \chi_{-m}(y) U(m)v\right\|^2 &
  = \bigg\langle \sum_{m \le N} \chi_{-m}(y) U(m)v, \sum_{k\le N} \chi_{-k}(y) U(k)v\bigg\rangle \\
& = \sum_{m\le N} \sum_{k \le N} \big\langle\chi_{-m}(y) U(m)v,\chi_{-k}(y) U(k)v\big\rangle \\
& = \sum_{m\le N} \sum_{k \le N} \chi_{k-m}(y)\big\langle U(-k)U(m)v, v\big\rangle \\
& = \sum_{m\le N} \sum_{k \le N}e^{2\pi i (k-m) \cdot y}\big\langle U(m-k)v, v\big\rangle \\
\end{align*}
Now look at the terms
$T_l = e^{2\pi i l \cdot y}\big\langle U(l)v, v\big\rangle$
that appear from the change of variables $l = m - k$.
The peak contribution to the sum comes from $l = 0$ (with $(2N + 1)^d$
terms),
and they decay as $\lvert l\rvert$ increases,
until when some coordinate becomes greater than $2N$, when it vanishes.
Therefore, the mean defining $d\mu_{v,N}(y)$ corresponds to a Cèsaro sum of the Fourier series
$$\sum_{l\le N} e^{2\pi i l \cdot y}\big\langle U(l)v, v\big\rangle.$$
Thus $d\mu_{v,N}(y)$ converges to a (tempered) distribution $T$.
A calculation shows that
$\langle T, \chi_j\rangle = \langle U(j) v, v\rangle$,
first integrating on the torus $\Bbb{T}^d$ to get only terms where $m-k =
j$,
and then taking the limit as $N\to\infty$.
Let's see why: the definition of $\langle T, \chi_j\rangle$ is the limit as $N\to\infty$ of
$$ \frac{1}{(2N+1)^d} \sum_{m\le N} \sum_{k \le N} \int_{\Bbb{T}^d}
 \chi_j (y) \chi_{k-m}(y) \big\langle U(m-k)v, v\big\rangle \, d\lambda(y).
$$
The orthogonality of the characters $\chi_j$ and $\chi_{m-k}$ over $\Bbb{T}^d$ reduce the sum to only
$$ \frac{1}{(2N+1)^d} \sum_{m,k\le N \, m-k = j} \big\langle U(m-k)v, v\big\rangle.
$$
Now, a volume argument shows that for each fixed $j$,
as $N\to\infty$, the number of solutions to $m-k = j$ with all coordinates of $m$ and $k$ smaller than $N$ is equal to $(2N+1)^d$ up to error terms of degree $d-1$ in $N$.
This establishes the functional calculus to the trigonometric polynomials,
with the uniform bound
$\lvert\langle T, P\rangle\rvert \leq \sup_{y\in \Bbb{T}^d} P(y)$.
With this bound, we can both extend the functional calculus
to all continuous funtions,
and by the Riesz-Markov theorem this distribution
is in fact a measure $\mu$ on $\Bbb{T}^d$.
To conclude, we note that the isometry between the spaces $L^2(\mu)$ and $H$
is given by the linear map
$$ \mathcal{U}: L^2(\mu) \to H\\
 P \mapsto P(U)v $$
constructed by the functional calculus above.
This is an isometry since
$$\langle P, Q\rangle_{L^2(\mu)} =
  \int_{\Bbb{T}^d} P(y) \bar{Q}(y) \, d\mu(y) =
  \langle P(U) Q(U)^* v, v\rangle_H =
  \langle Q(U)^* P(U) v, v\rangle_H =
  \langle P(U)v, Q(U)v\rangle_H. $$
(We used the commutativity of all $U(m)$ and $U^*(m)$, since they are unitary, to commute also $P(U)$ and $Q^*(U)$)
Since $v$ is cyclic, the function $\mathcal{U}$ is onto $H$,
so indeed we have an isomorphism of Hilbert spaces.
A: (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 $m\in\mathbb Z^d\cap\mathbb T^d$,
$$
\int_{\mathbb T^d}\chi_m\,d\mu_v=\mu_v(\{m\})=\langle U(m)v,v\rangle.
$$
