已知 $\cos \alpha=\frac{1}{3}, \cos (\beta-\alpha)=\frac{\sqrt{3}}{3}$, 且 $0<\beta<\alpha<\pi$, 则 $\cos \beta=$ ( )
$\because \cos \alpha=\frac{1}{3}, \cos (\beta-\alpha)=\frac{\sqrt{3}}{3}$, 且 $0<\beta<\alpha<\pi$,
    $\therefore-\pi<\beta-\alpha<0$,

$\therefore \sin \alpha=\sqrt{1-\frac{1}{9}}=\frac{2 \sqrt{2}}{3}, \sin (\beta-\alpha)=-\sqrt{1-\frac{1}{3}}=-\frac{\sqrt{6}}{3}, \therefore \cos \beta=\cos [(\beta-\alpha)+\alpha]$

$=\cos (\beta-\alpha) \cos \alpha-\sin (\beta-\alpha) \sin \alpha=\frac{1}{3} \times \frac{\sqrt{3}}{3}-\frac{2 \sqrt{2}}{3} \times\left(-\frac{\sqrt{6}}{3}\right)=\frac{5 \sqrt{3}}{9}$, 故选 D.