1. Given a positive real number $x$ satisfies $\lg ^{2015} x<\lg ^{2014} x<\lg ^{2016} x$, then the range of $x$ is
1. $\left(0, \frac{1}{10}\right)$.

Let $a=\lg x$, then $a \in \mathbf{R}, a^{2015}1$.
Thus, $a<-1$, i.e., $\lg x<-1$. Therefore, $0<x<\frac{1}{10}$.