14. 1木 * Let $x, y$ be positive real numbers, and satisfy:
$$
x \cdot y^{1+\lg x}=1 \text {. }
$$

Find the range of $x y$.
Taking the logarithm, we get $\lg x+(1+\lg x) \lg y=0$. Let $u=\lg x+\lg y$, then
$$
0=u+\lg x \lg y \leqslant u+\left(\frac{\lg x+\lg y}{2}\right)^{2}=u+\frac{u^{2}}{4} \text {. }
$$

Therefore, $u \leqslant-4$ or $u \geqslant 0$. Thus, $0<x y \leqslant 10^{-4}$ or $x y \geqslant 1$.
Hence, the range of $x y$ is $\left(0,10^{-4}\right] \cup[1,+\infty)$.