Let $G$ be a locally compact group and let $\mu$ be a left Haar measure. We know
that $\mu$ is unique up to a scalar in $\mathbf{R}_{>0}$. I don't know so much about unitary representations of groups but for the sake of convenience let us make the following definition:
Let $(V,\langle\ ,\ \rangle)$ be an Hilbert space over $\mathbf{C}$ of countable (finite or infinite) orthonormal Schauder basis. We let Let $GL_{cont}(V)$ be the set of bounded (with respect to the operator norm) linear operators on $V$. We may view $GL_{cont}(V)$ as a topological group via the discrete topology. Now let $M$ be a $\mathbf{C}$ vector space with a linear $G$-action. We will say that $M$ is a unitary irreducible representation of $G$ if there exists an abstract isomorphism of $\mathbf{C}$ vector spaces $f:M\rightarrow V$ (where $V$ is chosen as above) such that the natural map $GL(M)\rightarrow GL(V)$
(1) factors through $GL_{cont}(V)$
(2) $V$ is irreducible as a $G$-module
(3) For all $g\in G$ and all $v,w\in V$ one has that $\langle\rho(g)v,\rho(g)w\rangle=\langle v,w\rangle$.
Now let us consider the space $L^2(G)$ of all functions
$f:G\rightarrow\mathbf{C}$ where $f$ is measurable and square integrable with respect to the Haar measure. Note that this space has a natural structure of a $G$-module through left action.
Now in the special case where $G$ is a compact Lie group ($G$ is not necessarily connected so in particular this covers all finite groups) then all irreducible representation
are unitary (the average trick) and finite dimensional (this I think is non-trivial and follows from Peter-Weyl, actually I never looked at the proof of this result). Moreover, if
$\widehat{G}$ denote a complete set irreducible $\mathbf{C}$ representations of $G$ (up to isomorphisms as (unitary) $G$-modules) then one has that
$L^2(G)=\bigoplus_{\phi\in\widehat{G}}\oplus_{i,j}\sqrt{n_{\phi}}\phi_{ij}$ where $n_{\phi}=dim(\phi)$ and
$\phi_{ij}$ is the $(i,j)$-th entry of $\phi:G\rightarrow GL(V_{\phi})$. In other words all irreducible unitary representations (say $\phi$ is one of them) of $G$ occur in $L^2(G)$ with multiplicities $n_{\phi}$. The direct sum here should be understood in the sense of Schauder basis with respect to the topology induced by $\langle\ ,\ \rangle$. Note that
$\lbrace\sqrt{n_{\phi}}\phi_{i,j}\rbrace$ gives an orthonormal basis of $L^2(G)$.
Now here is a set of natural questions:
(1) Do all the irreducible unitary representations of a semi-semiple (reductive)
algebraic group over $\mathbf{R}$ occur in $L^2(G)$?
(2) On the other side of the spectrum, what about algebraic solvable groups?
(3) What is the minimal example of a locally compact topological group $G$ (with an non artificial tailor made topology, in particular $G$ has to be infinite) for which one can find an irreducible unitary representation which does not occur in $L^2(G)$?
Best Answer
I'm not quite sure if this is the answer that you looking for but anyway he we go. For a locally compact group you are going to generally want to look at strongly continuous representation. By this is mean endow $B(H)$, the bounded operators on a hilbert space $H$ with the topology of point-wise norm convergence. And only consider reps $\pi:G\rightarrow B(H)$ that are continuous with this topology. Now such a rep is unitary if, for every $g\in G$, $\pi(g)$ is a unitary operator.
Now the notion of "occur in" that you mention seems to be the notion of strong containment. We say that $\rho:G\rightarrow B(K)$ is strongly contained in $\pi:G\rightarrow B(H)$ if there is a $G$-equivarient unitary operator from $K$ to a closed subspace of $H$.
So it now seems that you are asking when does the left regular rep ($L^2(G)$) strongly contain all irreducibles. So yes for compact Lie groups this follows from Peter-weyl this is true.
However as soon as you go to something non-compact this might not be true.
In fact, there is a much weaker notion known as weak containment of representation. and it is known that $L^2(G)$ weakly contains all irreducible reps if and only if $G$ is amenable.
Non-compact Lie groups are in general not amenable, (any groups which contains $\mathbb{F}_2$ the free group on 2 generators is non-amenable)
There is much more to be said about this but I think that this should suffice for now