How to use SNAG theorem to find the unitary representations of the additive group $\mathbb{R}^n$? Consider the additive group $\mathbb{R}^n$. It is an abelian, locally compact group. The SNAG theorem applies. That theorem says:

SNAG Theorem: Let $T$ be an unitary continuous representation of an abelian locally compact group $G$ in a Hilbert space $\mathscr{H}$. Then there exists on the character group $\hat{G}$ a spectral measure $E$  such that $$T(x)=\int_{\hat{G}}\langle \hat{x},x\rangle dE(\hat{x})$$

It is said that for $\mathbb{R}^n$ there are $n$ strongly commuting self-adjoint operators $Y_1,\dots, Y_n$ with
$$T_x=\prod_{k=1}^n\exp(i x_k Y_k)$$
Now, how can this be derived? 
I think the starting point is to find the character group. Since a character is a continuous group homomorphism $\chi : G\to S^1$ it is $\chi(g)=e^{i\theta(g)}$ for $\theta : G\to \mathbb{R}$ a continuous function with $\theta(gh)=\theta(g)+\theta(h)$.
For the $G = \mathbb{R}^n$ the function $\theta$ is continuous and satisfies $\theta(g+h)=\theta(g)+\theta(h)$ so it is linear.
Hence fixing a basis $\{e_i\}$ of $\mathbb{R}^n$ with dual basis $\{\omega^i\}$ expanding $\theta = \sum \theta_i \omega^i$ and $g =\sum g^ie_i$ we have
$$T_g=\int_{(\mathbb{R}^n)^\ast}\exp(i\sum_k\theta_k g^k) dE(\theta)=\int_{(\mathbb{R}^n)^\ast}\prod \exp(i \theta_k g^k)dE(\theta).$$
I think now we would need to split the integral as a product of integrals and then identify $Y_i$. But how this is done? Why the integral can be split as a product? How the $Y_i$ are identified?
How, rigorously, the SNAG theorem gives the unitary representations of $\mathbb{R}^n$ in that form?
 A: Denote by $\pi_k$ the projection onto the $k$-th coordinate. If $E$ is a spectral measure on $\mathbb{R}^n$, then the operators $Y_i$ defined by
\begin{align*}
D(Y_k)&=\left\{\xi\in\mathscr{H} : \int \pi_k(\theta)^2\,d\langle E(\theta)\xi,\xi\rangle<\infty\right\},\\
Y_k\xi&=\int \pi_k(\theta)\,dE(\theta) \xi
\end{align*}
are self-adjoint with spectral measure $E_k=(\pi_k)_\#(E)$. In particular, these operators commute strongly.
To see that these operators yield the desired representation of $T$, first note that the spectral integral $f\mapsto \int f\,dE$ is a $\ast$-homomorphism from the algebra of bounded Borel functions to the algebra of bounded linear operator on $\mathscr{H}$: For simple functions $f=\sum_j a_j 1_{A_j}$ and $g=\sum_k b_k 1_{B_k}$ one has
$$
\int fg\,dE=\int \sum_{j,k}a_j b_k 1_{A_j\cap B_k}\,dE=\sum_{j,k}a_j b_k E(A_j)E( B_k)=\left(\sum_j a_j E(A_j)\right)\left(\sum_k b_k E(B_k)\right)=\left(\int f\,dE\right)\left(\int g\,dE\right).
$$
For arbitrary bounded Borel functions $f$, $g$ the result follows by continuity. Similarly one can verify linearity.
Thus
$$
\int \prod_{k=1}^n f_k(\theta_k)\,dE(\theta)=\prod_{k=1}^n \int f_k(\theta_k)\,dE(\theta)=\prod_{k=1}^n f_k(Y_k)
$$
for all bounded Borel functions $f_1,\dots,f_n$.
