Bahadur's theorem says that every bounded complete sufficient statistic is also minimal sufficient. But any minimal sufficient statistic is a one-to-one function of any other minimal sufficient statistic，which implies any minimal sufficient statistic is also a one-to-one function of a bounded complete sufficient statistic.
Thus if a bounded complete sufficient statistic exists, then every MSS is a one-to-one function of it, and thus every MSS is also complete.
Is this right? I feel it is wrong, but I don't know where the flaw is.
Indeed, any statistic based only on complete statistic is also complete. Suppose $T$ a complete statistic and $T^\prime=h(T)$ for some function $h$. Suppose for some measurable function $g$, $E_\theta(g(T^\prime))=0$ for all $\theta\in\Theta$. It follows that $E_\theta(g(h(T)))=0$ for all $\theta\in\Theta$, and thus $P_\theta(g(h(T))=0)=1$ and that is $P_\theta(g(T^\prime)=0)=1$. Therefore by definition $T^\prime$ is complete.
Since any minimal sufficient statistic by definition is a function of any sufficient statistic, thus if a complete sufficient statistic exists, that minimal sufficient statistic must be a function of it and is therefore complete sufficient too.