I have a task in optimization to do. That is \begin{align} \min_{z} \hat{J}(z):=\int_Q |u(z)-u_d| \end{align} such that \begin{align} &u_t-\Delta u=f\ \ \ \ on \ \ Q \\ &\partial u_n=0 \ \ \ \ on \ \Sigma \\ & u=z \ \ \ \ in \ \ \Omega \times \{t=0\} \end{align} where $$z \in \mathcal{Z}=\{z \in \mathcal{M(\bar{\Omega}): \|u\| \leq \alpha,\ \alpha>0}\}$$ and $Q=\Omega \times (0,T),\ \ \Sigma= \partial \Omega \times (0,T)$, $\partial \Omega$ refers to the boundary of $\Omega$. Also, $u(z)$ means the solution $u$ of the above pde with $z=u$ in $\Omega \times \{t=0\},$ $u_d$ is given and it is defined on $Q$, $f \in \left(H^1(Q)\right)^*$, i.e in the daul space of $H^1(Q)$ and $\partial_nu=\nabla u.n$ the normal derivative.
My question is how is that $u=z$ in $\Omega$? By definition of $\mathcal{M}(\bar{\Omega})$ that I google, $z \in \mathcal{M}(\bar{\Omega})$ means it is a measure that is defined on $B(\bar{\Omega})$ where the latter notation means the Borel field of $\Omega$ which consists of subsets of $\bar{\Omega}$, so logically $z$ is not defined on $x$ it is defined on $\{x\}$ which contradicts what I have in the above pde. So could anyone explain how the elements of $\mathcal{M}(\bar{\Omega})$, and how they are defined on $\Omega$?