If $(\rho,V)$ is a unitary representation of a group $G$ which is finite dimensional, then complete reducibility is kind of easy to prove. Indeed, if $V$ is not irreducible, then it has one proper invariant subspace $W$. The orthogonal complement of $W$, namely $W^\perp$ is then another proper invariant subspace so that $V$ decomposes as
$$V = W\oplus W^\perp,$$
decomposing $\rho$ as $\rho|_W\oplus \rho|_{W^{\perp}}$. Furthermore, since $V$ is finite dimensional and $W$ is a proper subspace, $\dim W,\dim W^\perp < \dim V$.
We can repeat this process with $W,W^\perp$ and so on, until we get irreducible representations, and we are assured this will end because each step lowers the dimension.
Now consider the Poincaré group $\mathbb{R}^4\rtimes SL(2,\mathbb{C})$. One looks for unitary representations of this group. It is then one theorem that there are no finite dimensional unitary representations. So they are all infinite dimensional.
The procedure above doesn't work. Each step won't lower the dimension. Still, physicists seem to still try to somehow get complete reducibility here. What they do, which is actually quite cryptic really is: let $(U,\mathscr{H})$ be a unitary representation in the Hilbert space $\mathscr{H}$. Consider a pure translation, namely, $U(a,1)$. This is $U(a,1)=e^{-ia^\mu P_\mu}$ for hermitian operators $P_\mu$. Then they "diagonalize" these operators with the improper basis $\Psi_{p,\sigma}$ so that
$$P_\mu \Psi_{p,\sigma}=p_\mu \Psi_{p,\sigma}.$$
Then one works on this basis, and notices that a pure $SL(2,\mathbb{C})$ transformation, $U(0,\Lambda)$ acting on $\Psi_{p,\sigma}$ is such that
$$P_\mu U(0,\Lambda)\Psi_{p,\sigma}=\sum_{\sigma'}C_{\sigma\sigma'}(p,\Lambda)\Psi_{\Lambda p,\sigma'}.$$
From this, Weinberg, e.g., says:
In general it may be possible by using suitable linear combinations of the $\Psi_{p,\sigma}$ to choose the $\sigma$ labels in such a way that the matrix $C_{\sigma'\sigma}(\Lambda,p)$ is block-diagonal; in other words, so that the $\Psi_{p,\sigma}$ with $\sigma$ within any one block by themselves furnish a representation of the inhomogeneous Lorentz group. It is natural to identify the states of a specific particle type with the components of a representation of the inhomogeneous Lorentz group which is irreducible, in the sense that it cannot be further decomposed in this way.
So it seems that all this procedure I described above is some sort of "complete reducibility under the hood". It seems more like "complete reducibility parametrized by $p$" actually. But I confess I don't get the point.
Actually I've read in this blog post that the matter is actually so subtle so a "direct integral" decomposition would be required.
So what is going on here? How is complete reducibility actually dealt with for the Poincare group? What is this "physicists procedure" Weinberg shows and comments all about?