$f = g$ in $H^{-1}$ implies $f = g$ a.e.?

I have two functions $f, g \in H^1_0$. But I could only show $|f - g|_{H^{-1}} = 0$. Does that imply $f = g$ almost everywhere? I think this should follow immediately since if $f = g$ in $H^{-1}$, then$$\langle f - g, h \rangle = 0 \forall h \in H^1_0.$$We could set $h := f - g$ and it follows that $|f - g|_{L^2}^2 = 0$. Am I correct?