Any character $L$ of $L^1(G)$ is given by $L(\phi)=\int_G \phi(x)f(x)dx$ for some bounded function $f$ The proof of theorem 7, chp. 3, p.60, of Lang's SL2 starts with

By measure theory, given a character $L\neq0$ of the algebra $L^1(G//K)$, there exists a bounded measurable function $f$ such that $$L(\phi)=\int_F\phi(x)f(x)\,dx, \qquad \forall\phi \in L^1(G//K)$$

$L^1(G//K)$ is the convolution algebra of bi-invariant functions of a locally compact Hausdorff group $G$ wrt to a compact subgroup $K$, but what bi-invariance is is irrelevant here (which is why I only put $L^1(G)$ in the title).
I've studied measure theory before, so I'd probably be fine with a mere more precise reference of the theorems involved, but any more thorough answer will be greatly appreciated.
 A: First, let me clarify the definitions. Let $G$ be a locally compact Hausdorff group with Haar measure $\mu$. Then $L^1(G)$ is a commutative Banach algebra when equipped with convolution. For a given subgroup $K$, a function $f$ on $G$ is called bi-invariant (with respect to $K$) if $f(k_1 g k_2) = f(g)$ for all $g \in G$ and $k_1,k_2 \in K$. We denote by $L^1(G//K)$ the space of integrable bi-invariant functions. One can verify that it is in fact a subalgebra of $L^1(G)$.
For a Banach algebra $A$ over $\mathbb{C}$, a character of $A$ is a morphism of algebras $L: A \to \mathbb{C}$. Explicitely, $L$ is a linear functional which satisfies $L(x y)=L(x) L(y)$ (it is multiplicative). Such a map is always continuous, so it is in fact a morphism of Banach algebras.

Now let us prove the claim. Let $L$ be a character of the algebra $L^1(G//K)$. In particular, $L$ is a bounded linear functional on $L^1(G//K)$, which is a subspace of $L^1(G)$. By the Hahn-Banach theorem, $L$ has an extension to a linear functional $L'$ on $L^1(G)$, which has the same norm. Thus $L'$ is in the dual of $L^1(G)$.
For a general $\sigma$-finite measure space $(X,\nu)$, it is well-known that the dual of $L^1(X, \nu)$ is isomorphic to $L^{\infty}(X, \nu)$. Indeed, an isomorphism is given by
$$
f \in L^{\infty}(X, \nu) \longmapsto \left( \phi \mapsto \int_X \phi \, f \; d\nu \right) \in (L^1(X, \nu))^*
$$
Assuming that the Haar measure on $G$ is $\sigma$-finite, this result implies that there exists a function $f \in L^{\infty}(G)$ such that
$$
L'(\phi) = \int_G \phi \, f \; d\mu
$$
for all $\phi \in L^1(G)$. By choosing a suitable representative we can always assume that $f$ is bounded. Since $L'$ is equal to $L$ on $L^1(G//K)$, the claim is proved.

One difficulty is that the Haar measure on a locally compact Hausdorff group is not necessarily $\sigma$-finite, and the aforementioned map $L^{\infty} \to (L^1)^*$ can fail to be surjective in this case. However, the Haar measure is $\sigma$-finite, for example, if $G$ has a countable number of connected components, which is true in many cases.
If $\sigma$-finiteness is not assumed, this can still be fixed by changing slightly the definition of $L^{\infty}$.  A very good reference for this is section 2.3 of Folland's Abstract Harmonic Analysis (2nd ed.). See also this post on MSE. With this new definition (which is the same as the standard one when the space is $\sigma$-finite), we always have the isomorphism $L^{\infty} \cong (L^1)^*$ and the rest of the proof carries over.
