I am confused in this proof about how we obtain $\int f(u) \, d\mu(u) = \int f(u)g(u) \, d\mu(u)$ and how Plancherels theorem has been applied in $(6.6)$.
Furthermore, I cannot understand how if $\mu$ were supported by a compact set of $0$ L.measure this would imply that $g(u)=0$ for almost all $u$.
Note: The following answer assumes knowledge about tempered distributions. If you do not know about tempered distributions, I can try to reformulate it, but I would prefer not to :)
Your assumption implies that $\hat{\mu}$ is an $L^2$ function.
By Plancherel's theorem, the Fourier transform $\mathcal{F} : L^2 \to L^2, f \mapsto \hat{f}$ is an isomorphism. Hence, there is some $g \in L^2$ with $\hat{g} = \hat{\mu}$.
But the $L^2$ Fourier-transform coincides with the Fourier transform for tempered distributions. Furthermore, the Fourier transform is injective on the set of tempered distributions. Hence, $g = \mu$ in the sense of tempered distributions, which means
$$ \int f \, d\mu = \int f(x) g(x) \, dx $$
for all Schwartz functions $f \in \mathcal{S}$.
Using a density argument, one can generalize this to $f \in L^2 \cap L^\infty \supset C_c$.
Now assume that the support $K$ of $\mu$ is a null-set. This implies $0 = \int f \,d\mu = \int f(x) g(x) \,dx$ for all $f \in C_c$ with ${\rm supp}(f) \cap K = \emptyset$.
Using standard arguments, this yields $g(x) = 0$ for almost all $x \in \Bbb{R} \setminus K$, so that $g$ vanishes almost everywhere, because $K$ is a null-set.
But this finally implies
$$ 0 = \int f(x) g(x) \,dx = \int f \,d\mu $$
for all $f \in C_c$ and hence $\mu \equiv 0$, which implies $\hat{\mu} \equiv 0$ in contradiction to the assumption of the theorem.
EDIT: Ok, so let's do it using only the $L^{2}$-theory of the Fourier transform (which is also used in the original proof, so this should be ok). I did not really know how much else I could use, so it got quite long:
By assumption, the function $\widehat{\mu}$ is in $L^{2}\left(\mathbb{R}\right)$. Plancherel's theorem states (in one formulation) that the Fourier transform $$ \mathcal{F}:L^{2}\left(\mathbb{R}\right)\to L^{2}\left(\mathbb{R}\right),f\mapsto\widehat{f} $$ is an (isometric) isomorphism. Hence, $f:=\mathcal{F}^{-1}\widehat{\mu}\in L^{2}\left(\mathbb{R}\right)$ is a well-defined $L^{2}$ function.
Now let $g\in L^{1}\left(\mathbb{R}\right)\cap L^{2}\left(\mathbb{R}\right)$ be arbitrary. Using Fubini's theorem, we have \begin{eqnarray*} \int_{\mathbb{R}}\overline{g\left(\xi\right)}\cdot\widehat{\mu}\left(\xi\right)\,{\rm d}\xi & = & \int_{\mathbb{R}}\int_{\mathbb{R}}\overline{g\left(\xi\right)}\cdot e^{-2\pi ix\xi}\,{\rm d}\mu\left(x\right)\,{\rm d}\xi\\ & = & \int_{\mathbb{R}}\int_{\mathbb{R}}\overline{g\left(\xi\right)}\cdot e^{-2\pi ix\xi}\,{\rm d}\xi\,{\rm d}\mu\left(x\right)\\ & = & \int_{\mathbb{R}}\overline{\int_{\mathbb{R}}g\left(\xi\right)\cdot e^{2\pi ix\xi}\,{\rm d}\xi}\,{\rm d}\mu\left(x\right)\\ & = & \int_{\mathbb{R}}\overline{\left(\mathcal{F}^{-1}g\right)\left(x\right)}\, d\mu\left(x\right), \end{eqnarray*} because the inverse Fourier transform is (on $L^{1}\cap L^{2}$) given by $$ \left(\mathcal{F}^{-1}\psi\right)\left(x\right)=\int_{\mathbb{R}}\psi\left(\xi\right)\cdot e^{2\pi ix\xi}\,{\rm d}\xi. $$
But Plancherel's theorem also implies (this is the isometry part) that $$ \left\langle \widehat{\varphi},\widehat{\psi}\right\rangle _{L^{2}}=\left\langle \varphi,\psi\right\rangle _{L^{2}}\qquad\forall\varphi,\psi\in L^{2}\left(\mathbb{R}\right). $$ If we apply this to the above, we also get (this is more or less equation (6.5)): \begin{eqnarray*} \int_{\mathbb{R}}\overline{\left(\mathcal{F}^{-1}g\right)\left(x\right)}\,{\rm d}\mu\left(x\right) & = & \int_{\mathbb{R}}\widehat{\mu}\left(\xi\right)\cdot\overline{g\left(\xi\right)}\,{\rm d}\xi\\ & = & \left\langle \widehat{\mu},g\right\rangle _{L^{2}}=\left\langle \mathcal{F}\mathcal{F}^{-1}\widehat{\mu},\mathcal{F}\mathcal{F}^{-1}g\right\rangle _{L^{2}}\\ & = & \left\langle \mathcal{F}^{-1}\widehat{\mu},\mathcal{F}^{-1}g\right\rangle _{L^{2}}\\ & = & \left\langle f,\mathcal{F}^{-1}g\right\rangle _{L^{2}}\\ & = & \int_{\mathbb{R}}f\left(x\right)\cdot\overline{\left(\mathcal{F}^{-1}g\right)\left(x\right)}\,{\rm d}x.\qquad\left(\dagger\right) \end{eqnarray*} Similarly (this is equation (6.6)), $$ \int_{\mathbb{R}}\left|f\left(x\right)\right|^{2}\,{\rm d}x=\left\langle f,f\right\rangle _{L^{2}}=\left\langle \mathcal{F}f,\mathcal{F}f\right\rangle _{L^{2}}=\left\langle \mathcal{F}\mathcal{F}^{-1}\widehat{\mu},\mathcal{F}\mathcal{F}^{-1}\widehat{\mu}\right\rangle _{L^{2}}=\left\langle \widehat{\mu},\widehat{\mu}\right\rangle _{L^{2}}=\int_{\mathbb{R}}\left|\widehat{\mu}\left(\xi\right)\right|^{2}\,{\rm d}\xi>0.\qquad\left(\ddagger\right) $$
Now let $h:\mathbb{R}\to\mathbb{R}$ be any continuous function vanishing outside some interval of finite length $\left[-R,R\right]$. One can then choose a sequence $\varphi_{n}$ of smooth, compactly supported functions with ${\rm supp}\left(\varphi_{n}\right)\subset\left[-\left(R+1\right),R+1\right]$ and with $\varphi_{n}\to h$ uniformly (for example using approximation by convolution, see this convergence of convolutions and approximation of unity for example). But \begin{eqnarray*} \left|\left(-2\pi i\xi\right)^{m}\cdot\widehat{\varphi_{n}}\left(\xi\right)\right| & = & \left|\int_{\mathbb{R}}\varphi_{n}\left(x\right)\cdot\left(-2\pi i\xi\right)^{m}e^{-2\pi ix\xi}\,{\rm d}x\right|\\ & = & \left|\int_{\mathbb{R}}\varphi_{n}\left(x\right)\cdot\frac{{\rm d}^{m}}{{\rm d}x^{m}}e^{-2\pi ix\xi}\,{\rm d}x\right|\\ & \overset{\text{partial integration}}{=} & \left|\left(-1\right)^{m}\cdot\int_{\mathbb{R}}\frac{{\rm d}^{m}}{{\rm d}x^{m}}\varphi_{n}\left(x\right)\cdot e^{-2\pi ix\xi}\,{\rm d}x\right|\\ & \leq & \int_{\mathbb{R}}\left|\frac{{\rm d}^{m}}{{\rm d}x^{m}}\varphi_{n}\left(x\right)\right|\,{\rm d}x=:C_{m,n}<\infty \end{eqnarray*} for arbitrary $m\in\mathbb{N}$ (philosophy: smoothness of $\varphi_{n}$ leads to decay of $\widehat{\varphi_{n}}$). This easily implies $\widehat{\varphi_{n}}\in L^{1}\cap L^{2}$.
Now apply $\left(\dagger\right)$ with $g=\widehat{\varphi_{n}}$ (which yields $\mathcal{F}^{-1}g=\mathcal{F}^{-1}\mathcal{F}\varphi_{n}=\varphi_{n}$) to conclude $$ \int_{\mathbb{R}}\overline{h\left(x\right)}\,{\rm d}\mu\left(x\right)\xleftarrow[n\to\infty]{}\int_{\mathbb{R}}\overline{\varphi_{n}\left(x\right)}\,{\rm d}\mu\left(x\right)=\int_{\mathbb{R}}f\left(x\right)\cdot\overline{\varphi_{n}\left(x\right)}\,{\rm d}x\xrightarrow[n\to\infty]{}\int_{\mathbb{R}}f\left(x\right)\cdot\overline{h\left(x\right)}\,{\rm d}x. $$ The convergence follows (e.g.) by dominated convergence.
Now assume that $N:={\rm supp}\left(\mu\right)\subset\left[-R,R\right]$ is a null-set. For each $n\in\mathbb{N}$ there is then an open set $U_{n}\supset N$ with $\lambda\left(U_{n}\right)<\varepsilon$ (where $\lambda$ is Lebesgue measure on the line). Hence $K_{n}:=\left[-R,R\right]\setminus U_{n}$ is a compact subset of the open set $\mathbb{R}\setminus N$, so that there is some continuous function $\gamma_{n}:\mathbb{R}\to\left[0,1\right]$ with $\gamma_{n}|_{K_{n}}\equiv1$ and $\gamma_{n}|_{N}\equiv0$. But then $\gamma_{n}\cdot h\xrightarrow[n\to\infty]{}h$ almost everywhere and $\left|\gamma_{n}\cdot h\right|\leq\left|h\right|$, so that dominated convergence yields $$ \int_{\mathbb{R}}f\left(x\right)\cdot\overline{h\left(x\right)}\,{\rm d}x\xleftarrow[n\to\infty]{}\int_{\mathbb{R}}f\left(x\right)\cdot\overline{\left(\gamma_{n}\cdot h\right)\left(x\right)}\,{\rm d}x=\int_{\mathbb{R}}\overline{\gamma_{n}\cdot h}\,{\rm d}\mu\overset{\gamma_{n}\cdot h\equiv0\text{ on }N={\rm supp}\left(\mu\right)}{=}0. $$ As this holds for all compactly supported continuous $h$, we get $f\equiv0$ almost everywhere, which contradicts $\left(\ddagger\right)$.