rt.representation theory - closed irreducible subspaces in L²(R)

Taking Fourier transforms, you get the action by multiplication. (Really, $x$ acts by multiplying with $e^{ix}$.) The closed invariant subspaces will thus be of the form ${fin L^2(mathbb{R})colonhat f|_E=0}$ for measurable sets $E$.

But you were asking for closed irreducible subspaces: There aren't any, because the Lebesgue measure has no atoms.