4. In tetrahedron $ABCD$, it is known that
$$
\angle ADB = \angle BDC = \angle CDA = \frac{\pi}{3} \text{, }
$$
the areas of $\triangle ADB$, $\triangle BDC$, and $\triangle CDA$ are $\frac{\sqrt{3}}{2}$, $2$, and $1$ respectively. Then the volume of this tetrahedron is $\qquad$
4. $\frac{2 \sqrt{6}}{9}$.

Let $D A, D B, D C$ be $x, y, z$ respectively. Then
$$
\frac{x y \sin \frac{\pi}{3}}{2}=\frac{\sqrt{3}}{2}, \frac{y z \sin \frac{\pi}{3}}{2}=2, \frac{x z \sin \frac{\pi}{3}}{2}=1 \text {. }
$$

Multiplying the three equations gives $x y z=\frac{8}{\sqrt{3}}$.
Let the angle between $D C$ and the plane $A B D$ be $\alpha$, and the distance from point $C$ to the plane $A B D$ be $h$. Then $h=z \sin \alpha$.
By the symmetry of the figure,
$\cos \frac{\pi}{3}=\cos \frac{\pi}{6} \cdot \cos \alpha \Rightarrow \cos \alpha=\frac{1}{\sqrt{3}}$
$\Rightarrow \sin \alpha=\frac{\sqrt{2}}{\sqrt{3}}$.
Therefore, the volume of the tetrahedron is
$\frac{1}{3} S_{\triangle A B D} \cdot h=\frac{1}{3}\left(\frac{x y \sin \frac{\pi}{3}}{2}\right) z \sin \alpha=\frac{2 \sqrt{6}}{9}$.