Why in an irreducible unitary representation of the Poincare group all momenta are on the same mass shell? This is a question about the approach of Weinberg in "The Quantum Theory of Fields" to the irreducible unitary representations of the Poincare group in Chapter 2.
Let $U(\Lambda,a)$ be such a representation. Weinberg shows that from it one is able to obtain the generator of translations $P^\mu$ acting on the Hilbert space. Weinberg picks $\Psi_{p,\sigma}$ its eigenstates, where $\sigma$ is a label to account for degeneracy.
The translation part acts trivially by a phase in this basis. Now if $U(\Lambda)$ is a homogeneous Lorentz transformation Weinberg shows that $$U(\Lambda)\Psi_{p,\sigma}=\sum_{\sigma'}C_{\sigma'\sigma}(\Lambda,p)\Psi_{\Lambda p,\sigma'},\tag{2.5.3}$$
so that $U(\Lambda)\Psi_{p,\sigma}$ is fully determined if one knows the coefficients $C_{\sigma'\sigma}(\Lambda,p)$.
Weinberg's approach to determine these is to write every momentum $p^\mu = L^\mu_{\phantom{\mu}\nu}(p)k^\nu$ in terms of a "standard $k^\nu$" which characterizes a class within which all $p^\mu$ has (1) the same $p^2=p_\mu p^\mu$ and (2) the same sign of $p^0$.
Here is my question: 

Why can he assume that all momenta in a irreducible unitary representation of the Poincare group have the same $p^2$ and the same sign of $p^0$? In other words, why all momenta in one irreducible representation of the Poincare group lie in the same mass shell?

My intuition says this has something to do with Schur's lemma. But I only know this result for finite-dimensional representations of finite groups and here we are dealing with infinite-dimensional unitary representations of a topological group.
 A: Both $p^2$ and the sign of $p^0$ are invariant under infinitesimal Poincaré transformations (we're really looking at representations of the Poincare algebra or rather projective representations of the connected component of the Poincaré group here).
Hence any representation that involves more than one value of $p^2$ and $\mathrm{sgn}(p^0)$ is not irreducible, as any subspace formed by the vectors with the same value for these quantities necessarily furnishes a subrepresentation of the Poincaré algebra.
A: Following the answer by @ACuriousMind I believe I got the point and decided to post a version which is a little bit more expanded with what I understood.
Let $f(p)$ be some function defined in momentum space. We can define the function acting on the momentum operators $P$ by the usual procedure using the eigenstate basis of the commuting set $\{P^\mu\}$:$$f(P)\Psi_{p,\sigma}=f(p)\Psi_{p,\sigma}.$$
Now take some $p_0$ and define $\mathcal{A}(p_0)$ to be the eigenspace of $f(P)$ with eigenvalue $f(p_0)$. This is spanned by all $\Psi_{p,\sigma}$ with $f(p)=f(p_0)$.
A general element of $\mathcal{A}(p_0)$ will be $$\Phi=\sum_{\sigma}\int_{f^{-1}(p_0)} {{\rm d}p \ } \Phi_\sigma(p) \Psi_{p,\sigma}.$$
Now act upon such $\Phi$ with a Lorentz transformation in the connected component with the identity $U(\Lambda)$:
$$U(\Lambda)\Phi=\sum_{\sigma}\int_{f^{-1}(p_0)} {{\rm d}p\ } \Phi_{\sigma}(p) U(\Lambda)\Psi_{p,\sigma}.$$
Now evaluate $f(P)$ of the above vector. As Weinberg shows, $U(\Lambda)\Psi_{p,\sigma}$ is eigenvector of $P^\mu$ with eigenvalue $\Lambda^\mu_{\nu}p^\nu$. In that case it is also eigenvector of $f(P^\mu)$ with eigenvalue $f(\Lambda^\mu_\nu p^\nu)$.
This means that $f(P)U(\Lambda)\Phi=f(\Lambda p)U(\Lambda)\Phi$ and hence the action of the Lorentz group, $U(\Lambda)$, takes $\mathcal{A}(p_0)$ to $\mathcal{A}(\Lambda p_0)$.
Now if $f(p)$ is a function such that $f(p)=f(\Lambda p)$ then all of the above shows that $U(\Lambda)$ takes $\mathcal{A}(p)$ into itself. In this regard, $\mathcal{A}(p)$ is one invariant subspace.
So if $f(P)$ has more than one eigenvalue it means that there is one proper invariant subspace and the representation is not irreducible. So in an irreducible representation all functions $f(p)$ which satisfy $f(p)=f(\Lambda p)$, i.e., which are invariant with respect to proper orthochronous Lorentz transformations, are such that $f(P)$ has only one eigenvalue. The only such two functions of $P$ are then $f(P) = P^2$ and $f(P) = \operatorname{sgn}P^0$.
