Let a (pure or mixed) quantum state of a one-dimensional particle be represented by a Wigner function $$W(x,p)$$. Here, $$x$$ and $$p$$ are the particle's position and momentum, respectively.
The density operator $$\rho$$ representing the state can be expressed in terms of the Wigner function: $\rho = \int_{-\infty}^{+\infty} dx \int_{-\infty}^{+\infty} dx' \int_{-\infty}^{+\infty} dp \, | x' \rangle \, e^{i p (x' - x) / \hbar} W \left( \frac{x + x'}{2}, p \right) \langle x | \,.$ Proof: According to the definition of the Wigner function, $W(x,p) = \frac{1}{2 \pi \hbar} \int_{-\infty}^{+\infty} dx' \, e^{-i p x' / \hbar} \langle x + \tfrac{1}{2} x' | \rho | x - \tfrac{1}{2} x' \rangle \,.$ Hence, \begin{align} \int_{-\infty}^{+\infty} dp \, e^{i p \xi' / \hbar} W(\xi,p) &= \frac{1}{2 \pi \hbar} \int_{-\infty}^{+\infty} d\xi'' \, \langle \xi + \tfrac{1}{2} \xi'' |\rho | \xi - \tfrac{1}{2} \xi'' \rangle \underbrace{ \int_{-\infty}^{+\infty} dp \, e^{i p (\xi' - \xi'') / \hbar} }_{2 \pi \hbar \delta(\xi' - \xi'')} \\ &= \langle \xi + \tfrac{1}{2} \xi' |\rho | \xi - \tfrac{1}{2} \xi' \rangle \,. \end{align} Introducing $$x = \xi - \tfrac{1}{2} \xi'$$ and $$x' = \xi + \tfrac{1}{2} \xi'$$, so that $$\xi = \tfrac{1}{2} (x + x')$$ and $$\xi' = x' - x$$, one gets $\langle x' | \rho | x \rangle = \int_{-\infty}^{+\infty} dp \, e^{i p (x' - x) / \hbar} W \left( \frac{x + x'}{2}, p \right) \,.$
Consider an observable represented by an operator $$A$$. The observable can be described by a phase-space function $$\mathcal{A}(x,p)$$ that is defined by requiring that its averaging over the Winger function yields the correct expectation value of the operator, i.e. $\operatorname{Tr} \rho A = \int_{-\infty}^{+\infty} dx \int_{-\infty}^{+\infty} dp \, W(x,p) \mathcal{A}(x,p) \,.$ It follows from this definition that $$\mathcal{A}(x,p)$$ is given by $\mathcal{A}(x,p) = \int_{-\infty}^{+\infty} dx' \, e^{-i p x' / \hbar} \langle x + \tfrac{1}{2} x' | A | x - \tfrac{1}{2} x' \rangle \,.$ Proof: The expression $\rho = \int_{-\infty}^{+\infty} dx \int_{-\infty}^{+\infty} dx' \int_{-\infty}^{+\infty} dp \, | x' \rangle \, e^{i p (x' - x) / \hbar} W \left( \frac{x + x'}{2}, p \right) \langle x |$ implies that \begin{align} \operatorname{Tr} \rho A &= \int_{-\infty}^{+\infty} d\xi'' \, \langle \xi'' | \rho A | \xi'' \rangle \\ &= \int_{-\infty}^{+\infty} d\xi \int_{-\infty}^{+\infty} d\xi' \int_{-\infty}^{+\infty} d\xi'' \int_{-\infty}^{+\infty} dp \, \langle \xi'' | \xi' \rangle \, e^{i p (\xi' - \xi) / \hbar} W \left( \frac{\xi + \xi'}{2}, p \right) \langle \xi | A | \xi'' \rangle \\ &= \int_{-\infty}^{+\infty} d\xi \int_{-\infty}^{+\infty} d\xi' \int_{-\infty}^{+\infty} dp \, e^{i p (\xi' - \xi) / \hbar} W \left( \frac{\xi + \xi'}{2}, p \right) \langle \xi | A | \xi' \rangle \end{align} \,. Introducing $$x = \tfrac{1}{2} (\xi + \xi')$$ and $$x' = \xi - \xi'$$, so that $$\xi = x + \tfrac{1}{2} x'$$ and $$\xi' = x - \tfrac{1}{2} x'$$, one gets $\operatorname{Tr} \rho A = \int_{-\infty}^{+\infty} dx \int_{-\infty}^{+\infty} dp \, W(x,p) \underbrace{ \int_{-\infty}^{+\infty} dx' \, e^{-i p x' / \hbar} \langle x + \tfrac{1}{2} x' | A | x - \tfrac{1}{2} x' \rangle }_{\mathcal{A}(x,p)} \,.$