The dual of $H^1_0(\Omega)$ is defined to $H^{-1}(\Omega)$. But what is the dual of $H^{-1}(\Omega)$? Is it $H^1_{0}(\Omega)$?
I am solving a problem which requires me to use the dual of $H^{-1}(\Omega)$ to compute a certain norm.
I am confused since $H^{-1}(\Omega)$ is not a 'natural' dual for $H^{1}_0(\Omega)$?
EDIT: I have a map $f:\mathbb{R}\to H^{-1}(\Omega)$, which I need to show measurable; and for nice domains this equivalent to weakly measurable, so I need to test measurability of every complex valued map given by $T\mapsto<f(T),g>_{H^{-1}(\Omega),H^{-1}(\Omega)^*}$.
Okay so (strong) measurability of $f$ is equivalent to weak measurability and $f$ being a.e. separably valued. Since $H^{-1}(\Omega)$ is separable we get the equivalence you mentioned.
Now to show weak measurability we have to show that for every $L\in H^{-1}(\Omega)^*$ the map $t\mapsto \langle f(t), L\rangle$ is measurable. So far this is exactly what you already have. To finish just recall that $H_0^1(\Omega)$ is a reflexive space, by virtue of being Hilbert, which means that $L=u^*$ for some $u\in H_0^1(\Omega)$ where $u^*(v)=v(u)$ for every $v\in H^{-1}(\Omega)$. Therefore it's enough to show that for every $u\in H^1_0(\Omega)$ we have $\langle f(t), u^*\rangle = \langle u, f(t)\rangle$ is measurable.
In other words, you don't need to worry about the structure of the dual space of $H^{-1}(\Omega)$.