The regular representation of R is not a Hilbert direct sum of irreducibles

Suppose $\lambda$ has a nonzero irreducible unitary subrepresentation on a closed invariant subspace $M\subseteq L^2(\mathbb{R})$; for $t\in \mathbb{R}$ the operator $\lambda (t){\mathrm{\mid }}_M$ is a bounded self-intertwiner of the restriction because $\lambda (t)\lambda (s)=\lambda (s)\lambda (t)$ for all $s,t$ (the group is abelian), so [F2] makes it a scalar $\chi (t)$ times the identity, and then every one-dimensional subspace of $M$ is invariant, so irreducibility [F3] forces $M=\mathbb{C}v$ with $\mathrm{\parallel }v\mathrm{\parallel }=1$ and $\lambda (t)v=\chi (t)v$ for all $t$; by [F6] this gives $\mathrm{\mid }v(x-t)\mathrm{\mid }=\mathrm{\mid }v(x)\mathrm{\mid }$ for every $t$ and almost every $x$, so $v=0$ almost everywhere by [F5], contradicting $\mathrm{\parallel }v\mathrm{\parallel }=1$; hence $\lambda$ has no nonzero irreducible unitary subrepresentation.

If $\lambda$ were a Hilbert direct sum of finite-dimensional irreducible unitary subrepresentations, then by [F4] at least one summand would be nonzero because $L^2(\mathbb{R})\mathrm{\ne }\{0\}$ by [F1], and that summand would be a nonzero irreducible unitary subrepresentation of $\lambda$, contradicting step 1.1; therefore no such decomposition exists, and the refuted statement fails. The Axiom of Choice enters through the cited Schur lemma, the Hilbert-direct-sum and regular-representation suppliers and the Lebesgue-measure facts of [F1] (the complete-measure and Radon-measure theorems are proved under the Axiom of Countable Choice); the vanishing argument itself is choice-free apart from those inputs.

In this scalar case the direct integral ${\int }_{\mathbb{R}}^{\oplus }\mathbb{C}d\xi$ means the Hilbert space of measurable scalar sections $h(\xi )$ with $\int \mathrm{\mid }h(\xi ){\mathrm{\mid }}^{2}d\xi <\mathrm{\infty }$, modulo null equality, with its integral inner product; this is exactly $L^2(\mathbb{R},d\xi )$. Let its fiberwise action be $(D(t)h)(\xi )=e^{-2\pi it\xi }h(\xi )$. Unit modulus makes each $D(t)$ unitary, the exponential addition law makes it a representation, and dominated convergence with majorant $4\mathrm{\mid }h{\mathrm{\mid }}^2$ proves strong continuity. On Schwartz inputs [F7] gives ${\mathcal{F}}_2\lambda (t)=D(t){\mathcal{F}}_2$; both sides are bounded operators, so density extends this identity to every $L^2$ class. The surjective unitary ${\mathcal{F}}_2$ therefore realizes $\lambda$ as the asserted direct integral of one-dimensional characters. For a fixed frequency the character has spatial modulus1, whose squared integral over $\mathbb{R}$ is infinite, so it is no nonzero vector in the original $L^2$ space. This explicit scalar integral supplies the motivating contrast without assuming general direct-integral decomposition or uniqueness theory.