Complete reducibility, in linear algebra versus topology. The example of the translation representation of $\mathbb R$ in $L^2({\mathbb{R}})$ On page 2 of these notes, the author provides an example of how a topologically reducible representation of a group can be lacking topological irreducibles. Before asking questions, I offer a simple introduction.
INTRODUCTION
Consider the representation of $\mathbb R$ on $L^2({\mathbb R})$ by translations $
 f(x)\;\longmapsto\;f(x\,+\,a)
 $
written as
$$
 {\mathbb{R}} \ni a\;:\quad f\;\longmapsto\;f\circ\hat{\cal{R}}_a\;\;.
 $$
An ${L^2({\mathbb R})}$ function can be Fourier-expanded as
$$
 f(x)\,=\int e^{\, 2\,\pi\, i\, t\, x}\, \hat{f}(t)\,dt\;\;.\qquad\qquad\qquad (*)
 $$
Since $\,f \in {L^2({\mathbb R})}\,\Longrightarrow\,\hat{\cal{R}}_a\, f \in {L^2({\mathbb R})}\,$, we can expand also the function $f\circ\hat{\cal{R}}_a\, (x) =f(x+a)\,$:
$$
 f(x\,+\, a)\,=\,\int e^{\, 2\,\pi\, i\, t\, x}\, \hat{f}_a(t)\,dt\;\;.
 $$
On the other hand, the insertion of $\, x+a\,$ instead of $\, x\,$ in (*) renders us
$$
 f(x+a)\,=\int e^{\, 2\,\pi\, i\, t\, x}\, e^{\, 2\,\pi\, i\, t\, a}\, \hat{f}(t)\,dt\;\;.
 $$
The former and the latter formulae, together, are indicating that the Fourier transform establishes an isomorphism between the translation representation of $\mathbb R$ on $L^2({\mathbb{R}})$ and the representation realised by multiplication by exponentials:
$$
 f(x)\,\longmapsto\,f(x\,+\, a)\quad\Longleftrightarrow\quad \hat{f}(t)\,\longmapsto\,e^{\, 2\,\pi\, i\, t\, a}\, \hat{f}(t)\;\;,\qquad\qquad\qquad (**)
 $$
so the Fourier transforms of $f$ and $f\circ\hat{\cal{R}}_a$ have the same support.
By Schur's lemma, an irreducible, if it exists, must be one-dimensional. This means that there is such a $t$ that (*) becomes
$$
 f(x)\,= e^{\, 2\,\pi\, i\, t\, x}\, \hat{f}(t)\;\;.
 $$
while (**) becomes
$$
 f(x+a)\,= e^{\, 2\,\pi\, i\, t\, x}\, e^{\, 2\,\pi\, i\, t\, a}\, \hat{f}(t)\;\;.
 $$
Combined, these two expressions yield:
$$
 f(x+a)\,= \, e^{\, 2\,\pi\, i\, t\, a}\,f(x)\;\;.\qquad\qquad\qquad (***)
 $$
Such a function, however, will not be residing in $L^2({\mathbb{R }})$.
Indeed, equality (***) enables us to write the squared norm $\,||\,f\,||^2\,=\,\int dx\,|f(x)|^2$ as
$$
 ||\,f\,||^2\,=\,\int da\,|\,e^{\, 2\,\pi\, i\, t\, a}\,f(0)\,|^2\,=\,|\,f(0)\,|^2\,\int da\,|\,e^{\, 2\,\pi\, i\, t\, a}\,|^2\,=\;\infty\;\;.
 $$
So the representation of $\mathbb R$ on $L^2({\mathbb{R}})$ lacks topological irreducibles.
At the same time, this representation is topologically reducible because it is said in these notes that, for any measurable set $S\subset\mathbb R$, the vector space of functions $f$ for which $\hat{f}$ has support in $S$ gives an invariant closed subspace.
QUESTION 1.
Is there an elementary proof of the said fact that, for any measurable $S\subset\mathbb R$, the space of functions $f$ for which $\hat{f}$ has support in $S$ is an invariant closed subspace?
Since $f$ and $f\circ{\cal{R}}_a$ have the same support, the invariance holds, and the issue is closedness. To prove it, can we somehow use the fact that the Fourier transform is an isometry?
QUESTION 2.
Would it be right to say that the representation is completely reducible algebraically, but not topologically?
To put it simply, can we say that the sole exponentials, not being topological irreducibles, still can be regarded as algebraic irreducibles?
QUESTION 3.
Below come two definitions of complete reducibility.  Equivalent in linear algebra and in finite-dimensional topological vector spaces, they are not necessarily equivalent in infinite dimensions.
Consider a representation A(G) of a group G is a TVS (topological vector space).
$\quad$ Def 1
$\quad$ A(G) is topologically completely reducible if it splits into a (finite or infinite) direct sum of topological irreducibles.
$\quad$ Def 2
$\quad$ A(G) is topologically completely reducible if any of its topological subrepresentations has a complementary topological subrepresentation.
The representation of $\mathbb{R}$ on $L^2({\mathbb{R}})$ evidently does not satisfy Def 1.
(This representation simply lacks topological irreducibles.)
Since for any measurable $S\subset\mathbb R$ the space of functions $f$ for which $\hat{f}$ has support in $S$ is an invariant closed subspace, may we claim that the representation of $\mathbb{R}$ on $L^2({\mathbb{R}})$ satisfies Def 2  (and is completely reducible in the sense of this Def)?  What makes me doubt this is the presence of the word `any' in that definition. By considering all measurable $S\subset\mathbb R$, will we count all topological subrepresentations?
 A: Throughout, I'll write just $\operatorname L^2$ for $\operatorname L^2(\mathbb R)$; and, when I refer to it as a representation, will always mean to think of it as a representation of $\mathbb R$, with the regular (translation) action.
QUESTION 1.  Since the Fourier transform is a unitary automorphism of $\operatorname L^2$, it suffices to show that the space of functions vanishing on $S$ is closed in $\operatorname L^2$.  A sequence that converges in $\operatorname L^2$ has a subsequence that converges pointwise almost everywhere; so, if the original sequence consists of functions vanishing almost everywhere on $S$, then the limit function vanishes almost everywhere on $S$.
QUESTION 2.  $\operatorname L^2$ is neither algebraically nor topologically completely reducible, in the sense of Definition 1.  There are no irreducible subrepresentations of $\operatorname L^2(\mathbb R)$, algebraic or topological, because they would have to be 1-dimensional, hence spanned by a function $f$; and that function $f$ would have to transform under translation by a (necessarily unitary) character $\chi$, hence, if non-$0$, could not lie in $\operatorname L^2$.  As in our discussion at MathOverflow, we would find that
$$
\int_{\mathbb R} \lvert f(t)\rvert^2\mathrm dt = \int_{\mathbb R} \lvert\chi(t - t_0) f(t_0)\rvert^2\mathrm dt = \int_{\mathbb R} \lvert f(t_0)\rvert^2\mathrm dt = \infty
$$
for any $t_0 \in \mathbb R$ such that $f(t_0) \ne 0$.
QUESTION 3:  I do not know (though I think it is known) whether all subrepresentations of $\operatorname L^2$ arise in this way.
