$\int_{\Omega} f \psi = 0$ for every $\psi$ which is zero at $\partial \Omega$ imples $f=0$ This is a generalization of my previous question. Let $I=[t_{0},t_{1}]\subset \mathbb{R}$ be fixed and  $\Omega \subset \mathbb{R}^{n}$. Let $\psi$ and $f$ be sufficiently differentiable (we can assume it to be smooth) and let us assume that $\psi(t_{0},x)= \psi(t_{1},x) = 0$ for every $x \in \mathbb{R}^{n}$ and $\psi(t,x)|_{\partial \Omega} = 0$ for every $t \in I$. If:
$$\int_{I}\int_{\Omega}f(t,x)\psi(t,x)dxdt = 0$$
for every $\psi$ satisfying the above conditions, is it true that $f(t,x) \equiv 0$?
 A: Consider just those $\psi(t,x) = \phi(t)g(x)$.
$$
0=\int_I \int_\Omega f(t,x)\psi(t,x)\,dx\,dt = \int_I \phi(t) \left(\int_\Omega f(t,x)g(x)\,dx\right)\,dt = \int_I \phi(t) F_g(t)\,dt.
$$
where $F_g$ is the inner integral.
The assumptions imply (e.g. by your previous question) that $F_g(t) = 0$ for all $t \in I$.
Fix $t \in I$ and let $f_t(x) := f(t,x)$.
Since $g$ (smooth enough etc.)  was unrestrained, we have
$$
0 = \int_\Omega f_t(x) g(x)\,dx
$$
for all smooth $g$.
Assume for sake of contradiction that $f_t(x_0) \neq 0$ for some $x_0$ on the interior of $\Omega$.
Take $g$ to be a positive smooth approximation of $1_{B(x_0)}$ where $B(x_0)$ is a ball around an arbitrary point $x_0$ such that $B(x_0) \subseteq \Omega$ and $f$ does not change sign in $B(x_0)$. Then we find that $\int_{B(x_0)}f_t(x) g(x) \, dx = 0$ but the integral of a smooth function of constant sign being zero means the function is zero, so we arrive at a contradiction, $f_t(x_0)$ must be $0$.
Hence $f_t(x_0) = 0$ for all $t \in I$ and all $x$ on the interior of $\Omega$.
Since for each $t$, $f_t$ is smooth, its value on the boundary of $\Omega$ is the limit of its values on the interior (here is where we assume $\Omega$ is not a pathological set), and hence $f_t(x) =0$ for all $t \in I, x \in \Omega$.
