Suppose $\mu$ is a finite signed Borel measure on $\mathbb{R}$. (Let's say it's real for convenience.) Suppose in addition that $\hat{\mu}$ (its Fourier transform) is an $L^{1}(\mathbb{R})$ function. I claim that if $E$ is a Borel set, then $\mu(E) = \int_{E} f(x) \, dx$, where $f(x) = \int_{\mathbb{R}} \hat{\mu}(\xi) e^{i 2 \pi \xi} \, d\xi$. I'll show this by proving that \begin{equation*} \forall \varphi \in \mathcal{S}(\mathbb{R}) \quad \int_{\mathbb{R}} \varphi(x) \, \mu(dx) = \int_{\mathbb{R}} \varphi(x) f(x) \, dx, \end{equation*} where $\mathcal{S}(\mathbb{R})$ is the Schwarz space.
The first thing to ascertain is whether or not $\hat{f}(\xi) = \hat{\mu}(\xi)$ on $\mathbb{R}$. However, this is clearly true in the sense of distributions so I won't attend to it further. (Note that $f \in L^{\infty}(\mathbb{R})$ is clear, but it's not immediately obvious to me at this point that $f \in L^{1}(\mathbb{R})$.)
Suppose $\varphi \in \mathcal{S}(\mathbb{R})$. Then $\varphi(x) = \int_{\mathbb{R}} \hat{\varphi}(\xi) e^{i 2 \pi \xi x} \, d\xi$ and, thus, \begin{align*} \int_{\mathbb{R}} \varphi(x) \, \mu(dx) &= \int_{\mathbb{R}} \int_{\mathbb{R}} \hat{\varphi}(\xi) e^{i 2 \pi \xi x} \, d \xi \mu(dx) \\ &= \int_{\mathbb{R}} \hat{\varphi}(\xi) \int_{\mathbb{R}} e^{i 2 \pi \xi} \mu(dx) d \xi \\ &= \int_{\mathbb{R}} \hat{\varphi}(\xi) \hat{\mu}(-\xi) \, d \xi \\ &= \int_{\mathbb{R}} f(x) \varphi(x) \, dx. \end{align*} Above I used the fact that $\hat{f} = \hat{\mu}$ in the sense of distributions. Since $\mathcal{S}(\mathbb{R})$ is dense in $C_{0}(\mathbb{R})$ --- indeed, the former contains $C_{c}^{\infty}(\mathbb{R})$---, it follows that $\mu = f \, dx$.
EDIT: I never completely addressed the question. Suppose now that $\varphi \in BUC(\mathbb{R}) \cap L^{1}(\mathbb{R})$ (BUC = bounded uniformly continuous functions) is Hermetian (meaning $\varphi(-\xi) = \overline{\varphi(\xi)}$), positive definite, and $|\varphi(x)| \leq \varphi(0) = 1$. The arguments presented above show that if I define $f \in L^{\infty}(\mathbb{R}; \mathbb{C})$ by $f(x) = \int_{\mathbb{R}} \varphi(\xi) e^{i 2 \pi \xi} \, d \xi$, then $\hat{f} = \varphi$ provided I can prove $f \in L^{1}(\mathbb{R})$. I claim that $f \geq 0$ and $\int_{\mathbb{R}} f(x) = 1$.
First, observe that $f(x) \in \mathbb{R}$ a.e. This follows from the Hermetian symmetry of $\varphi$. Also observe that $1 = \varphi(0) = \int_{\mathbb{R}} f(x) \, dx$. It remains to show $f \geq 0$. We do this by showing that if $\psi \in \mathcal{S}(\mathbb{R})$, then $\int_{\mathbb{R}} \psi(x)^{2} f(x) \, dx \geq 0$. Indeed, given such a $\psi$, we can write $\psi(x) = \int_{\mathbb{R}} \hat{\psi}(\xi) e^{i 2 \pi \xi x} \, d\xi$ and the integral converges uniformly (w.r.t. $x$). Therefore, if we write $\psi_{n}(x) = N^{-1} \sum_{j = -N^{2}}^{N^{2}} \hat{\psi}(jN^{-1}) e^{i 2 \pi jN^{-1} x}$, then $\psi_{n} \to \psi$ uniformly on $\mathbb{R}$. Integrating, we appeal to positive definiteness to find \begin{align*} \int_{\mathbb{R}} |\psi_{n}(x)|^{2} f(x) \, dx &= N^{-2} \sum_{i,j = -N^{2}}^{N^{2}} \hat{\psi}(iN^{-1}) \overline{\hat{\psi}(jN^{-1})} \int_{\mathbb{R}} e^{-i 2 \pi (i - j)N^{-1} x} f(x) \, dx \\ &= N^{-2} \sum_{i, j = -N^{2}}^{N^{2}} \hat{\psi}(iN^{-1}) \overline{\hat{\psi}(jN^{-1})} \varphi(iN^{-1} - jN^{-1}) \geq 0. \end{align*} Sending $n \to \infty$, we find \begin{align*} \int_{\mathbb{R}} \psi(x)^{2} f(x) \, dx \geq 0. \end{align*} Since this is true when $\psi \in \mathcal{S}(\mathbb{R})$, it follows that it's true when $\psi \in L^{1}(\mathbb{R})$ by density (and boundedness of $f$). Finally, observe that if $g \in C_{c}(\mathbb{R})$ and $g \geq 0$, then $\sqrt{g} \in C_{c}(\mathbb{R})$ and, thus, $\int_{\mathbb{R}} g(x) f(x) \, dx \geq 0$. Since this is true independently of $g$, it follows that $f \geq 0$. This completes the proof of Bochner's Theorem.
