Splitting the representation of $\mathbb{R}$ on $L^2(\mathbb{R})$ into irreducibles. On p2 of these notes, the author writes that the translation action of $\mathbb{R}$ on $L^2(\mathbb{R})$ is a reducible representation which does not contain any irreducible subrepresentations. Presumably by $L^2(\mathbb{R})$ he means the square integrable functions $f : \mathbb{R}\rightarrow\mathbb{C}$.
"Indeed, for any measurable set $S\subset\mathbb{R}$, the space of functions $f$ for which the Fourier transform $\hat{f}$ has support in $S$ is a proper closed invariant subspace."
This space is certainly invariant - how would one see that it is closed?
"By Schur's lemma, any irreducible subrepresentation $W\subset L^2(\mathbb{R})$ would have dimension 1, ie $W = \langle F\rangle$ where $F: \mathbb{R}\rightarrow\mathbb{C}$ is a function such that $F(x+r)$ is proportional to $F(x)$. Thus, $F(x)$ is proportional to $e^{\lambda x}$ for some $\lambda\in\mathbb{C}$, but no such function lies in $L^2$"
No issues with this.
"However, a representation "decomposes" into a (possibly uncountable) number of irreducibles. This is the theory of "unitary disintegration""
This is my main question - what does he mean by this? Googling "unitary disintegration" did not yield any relevant results. Furthermore, how can a representation decompose into irreducibles, but have no irreducible subrepresentations?
References would be appreciated.
 A: Since the Fourier transform is an isometry, your only question is: why the subspace of $L^2(\mathbb{R})$ of functions that are zero on a measurable subset $A$ of $\mathbb{R}$ is a closed subspace of $L^2(\mathbb{R})$. You can express the condition as
$$(\int_A |f^2(t)| dt)^{1/2} = 0$$ 
The Fourier transform realizes an isomorphism between the translation representation of $\mathbb{R}$ on $L^2(\mathbb{R})$ and the representation given by multiplication by exponentials. So the Fourier transform diagonalizes the regular representation. This happens for every group ( think how one diagonalizes circular matrices). 
Back to our representation of $\mathbb{R}$ on $L^2(\mathbb{R})$ by 
$$(s, f(t)) \mapsto e^{i st} \cdot f(t)$$
Since for every $A\subset \mathbb{R}$ so that $\mu(A)$, $\mu(\mathbb{R}\backslash(A)) > 0$ we can write a unitary decomposition
$$L^{2}(\mathbb{R})=L^2(A) \oplus L^2(\mathbb{R}\backslash A)$$ we may think it's enough to decompose $\mathbb{R}$ into ultimate measurable pieces and have a decomposition into irreducible. The problem is that $\mathbb{R}$ does not decompose into atoms. We can still write this representation as an integral of $1$-dimensional representations. But in our case, it is exactly considering a function as composed of all its values ( more or less). The integral of representations formalizes this thing. 
A: Let $G = \mathbb R$, and $V = L^2(\mathbb R)$, which is a unitary representation of $G$ with the translation action.  It is true (more generally of unitary representations of locally compact groups) that if $W \subset V$ is a subrepresentation, then there exists another subrepresentation $W'$ of $V$ such that $V = W \oplus W'$.  Consequently, $V$ is semisimple as an abstract representation of $G$: there exist irreducible subrepresentations $W_i$ of $V$ such that
$$V = \bigoplus\limits_i W_i.$$
This is an algebraic direct sum, which means for every $v \in V$, there exist unique $w_i \in W_i$, all but finitely many of which are $0$, such that $v = \sum\limits w_i$.
This is not what representation theorists who study Hilbert space representations are interested in.  They are interested, in decomposing representations into a direct sum of irreducibles in a different sense: that is, in the best case scenario they are interested in finding closed irreducible representations $W_i$ of $V$ (that is, closed subspaces $W_i$ of $V$ which are subrepresentations of $G$, and which themselves admit no nonzero, proper, closed  subrepresentations $W_i'$ of $W_i$) such that the $W_i$ are mutually orthogonal, and such that
$$V = \hat{\bigoplus\limits_i} W_i \tag{Hilbert space direct sum}$$
This last condition means that for every $v \in V$, there exist $w_i \in W_i$ such that $v = \sum\limits_i w_i$ (as an absolutely convergent sum in the norm topology).
This turns out to be impossible in this case, since $V = L^2(\mathbb R)$ does not have any closed, irreducible subrepresentations: if $W$ were such a subrepresentation, it would have to be one-dimensional and correspond to a unitary character of $\mathbb R$.  Then there would have to exist a function $f: \mathbb R \rightarrow \mathbb C$ and a real number $t$, such that
$$\int\limits_{\mathbb R} |f(x)|^2 dx < \infty$$
$$ f(x+y) = e^{ 2 \pi i ty} f(x) \tag{$x,y \in \mathbb R$}$$
and this is impossible.
