4. The number of elements in the set $\left\{(x, y) \left\lvert\, \lg \left(x^{3}+\frac{1}{3} y^{3}+\frac{1}{9}\right)=\lg x+\lg y\right.\right\}$ is
(A) 0
(B) 1
(C) 2
(D) More than 2
(B)
4. 【Analysis and Solution】 $\lg \left(x^{3}+\frac{1}{3} y^{3}+\frac{1}{9}\right)=\lg x+\lg y \Leftrightarrow x^{3}+\frac{1}{3} y^{3}+\frac{1}{9}=x y, x>0, y>0$. By the AM-GM inequality,
$$
x^{3}+\frac{1}{3} y^{3}+\frac{1}{9} \geqslant 3 \sqrt{x^{3}\left(\frac{1}{3} y^{3}\right) \cdot \frac{1}{9}}=x y \text {. }
$$

Equality holds if and only if $\left\{\begin{array}{l}x^{3}=\frac{1}{9} \\ \frac{1}{3} y^{3}=\frac{1}{9}\end{array}\right.$, solving these two equations yields
$$
x=\sqrt[3]{\frac{1}{9}}, y=\sqrt[3]{\frac{1}{3}}
$$

Therefore, the set contains only one point, which is $\left(\sqrt[3]{\frac{1}{9}}, \sqrt[3]{\frac{1}{3}}\right)$.