Suppose $\nu$ is a real measure on $(X,\mathcal{S})$. Define $\mu:\mathcal{S} \rightarrow [0,\infty)$ by
$\mu(E)=|\nu(E)|$ (measure of $\nu(E)$ ) We need to show that $\nu$ is positive measure on $(X,\mathcal{S})$ if and only if the range of $\nu$ is contained in $[0,\infty)$ or the range of $\nu$ is contained in $(-\infty,0]$.
I think the first direction is clear, as if $\nu$ is a positive measure then the range must be contained in $[0,\infty)$ ( not the negatives, right!!) but, how can we assume that $\nu$ is positive measure if the range of the set function is completely positive or completely negative.
By Hahn decomposition theorem we can write $\nu=\nu_1+\nu_2$ and $\nu_1(A)=\nu(A\cap E), \nu_2(E)=\nu(A \cap E^{c})$ for all $A$ (where $E$ is suitable measurable set). Here $E$ in a positive set and $E^{c}$ is a negative set for $\nu$. Now $\mu(X)=\mu(E)+\mu (E^{c})$ which gives $|\nu(X)|=|\nu(E)|+|\nu(E)|$. Hence $|\nu(E)+\nu(E^{c}|=|\nu(E)|+|\nu(E)|$. This implies that $\nu (E)$ and $\nu (E^{c})$ have the same sigm. But $\nu (E) \geq 0$ and $\nu (E^{c}) \leq 0$. Thus either $\nu (E)=0$ in which case $\nu_1 $ is the zero measure and $\nu=\nu_2 $ takes values in $(-\infty, 0]$ or $\nu (E^{c})=0$ in which case $\nu_2 $ is the zero measure and $\nu=\nu_1 $ takes values in $[0,-\infty)$. The converse part, as you alredy know is obvious.