I've tried to find the unitary irreducible representations of the additive group $(\mathbb{R},+)$ and came up with a pair of results, which I want to verify if are correct. They are:
Theorem: Let the additive group $G=(\mathbb{R},+)$ be given. Then any unitary representation $U : G\to \mathfrak{U}(\mathscr{H})$ acting on the Hilbert space $\mathscr{H}$ is isomorphic to a unitary representation $U^\diamond : G\to \mathfrak{U}(L^2(X,d\mu))$ for some measure space $(X,\mu)$ given by $[U^\diamond(t)\Psi](x)=e^{-ita(x)}\Psi(x)$ for a real $a\in L^2(X,d\mu)$.
Proof: Since a unitary representation is given by a continuous homomorphism $U$ on the strong topology, this means that $U$ is actually a strongly continuous $1$-parameter group of unitary operators, after all, $U(t+s)=U(t)U(s)$.
Hence Stone's theorem guarantees that there is one hermitian operator $A$ in $\mathscr{H}$ such that $U(t) = e^{-itA}$. This in turn can be rigorously characterized by the spectral theorem.
Since $A$ is hermitian, there is a measure space $(X,\mu)$ and a unitary isomorphism $\chi : \mathscr{H}\to L^2(X,d\mu)$ such that if we define $A^\diamond = \chi A\chi^\dagger$ then $A^\diamond$ is a multiplication operator in the sense that there is $a\in L^2(X,d\mu)$ which is actualy real, satisfying
$$[A^\diamond\Psi](x)=a(x)\Psi(x),\quad \forall x\in X.$$
In that case we can perfectly define $e^{-itA^\diamond}$ as
$$[e^{-itA^\diamond}\Psi](x)=e^{-ita(x)}\Psi(x),\quad \forall x\in X.$$
Hence we may as well define
$$e^{-itA}=\chi^\dagger e^{-itA^\diamond}\chi.$$
But here $U^\diamond(t)=e^{-itA^\diamond}$ provides a unitary representation of $G$ on $L^2(X,d\mu)$ and so the last equation shows that $\chi$ becomes a representation isomorphism between it and the original one, completing the proof.
Corolary: Let the additive group $G = (\mathbb{R},+)$ be given. Then the irreducible representations are parametrized by $\lambda \in \mathbb{R}$ and given by $U^\diamond : G\to \mathfrak{U}(L^2(X,d\mu))$ given by $U^\diamond(t)=e^{-i\lambda t}\mathbf{1}$ where $\mathbf{1}$ is the identity operator and $(X,\mu)$ is a measure space.
Proof: Let an irreducible representation $U : G\to \mathfrak{U}(\mathscr{H})$ be given. By the previous theorem it is isomorphic to a representation $U^\diamond : G\to \mathfrak{U}(L^2(X,d\mu))$ characterized by a real function $a\in L^2(X,d\mu)$ so that $U^\diamond(t)= e^{-itA^\diamond}$ for the associate multiplication operator $A^\diamond$. It should be obvious that $A^\diamond$ commutes with $U^\diamond(t)$ for every $t$. Hence this means
$$[U^\diamond(t),A^\diamond]=U^\diamond(t)\circ A^\diamond - A^\diamond \circ U^\diamond(t) = 0.$$
Hence by Schur's lemma, since the representation is irreducible, $A^\diamond$ acts by a multiple of the identity. Thus $a(x) = \lambda$ for some $\lambda \in \mathbb{R}$ and for all $x\in X$. In turn $U^\diamond(t) = e^{-i\lambda t}\mathbf{1}$ completing the proof.
Now I'm unsure if this is correct (I'm actually a bit new to functional analysis). Is the result correct, or have I made some remarkably wrong mistake?
Finaly, I feel strange that one arbitrary measure space $(X,\mu)$ is left lurking around. Wasn't it required to have a fully determined $(X,\mu)$ in the irreducible representations?