Show that $C^\infty(0,T; L^2(\Omega(t)))$ is dense in $L^2(0,T; L^2(\Omega(t)))$. Let $T > 0$, $\Omega \subset \mathbb R^2$ and $f:[0,T] \to \mathbb R^2$ a continuous and bounded function. We define
$$\Omega(t) = \Omega + f(t),$$
and $\widetilde \Omega \subset \mathbb R^2$ such that $\Omega(t)\subset \widetilde \Omega$ for all $t$. Moreover, we set
\begin{align}
L^2(0,T; L^2(\Omega(t))) &= \{v \in L^2(0,T; L^2(\widetilde \Omega ))~|~ v(t, \cdot)|_{\widetilde \Omega \backslash\Omega(t)} = 0 ~\text{ a.e. in }[0,T] \},\\
C^\infty(0,T; L^2(\Omega(t))) &= \{v \in C^\infty(0,T; L^2(\widetilde \Omega ))~|~v(t, \cdot)|_{\widetilde \Omega \backslash\Omega(t)} = 0~~\forall t \in [0,T] \}.
\end{align}
It is not too hard to show that $ C^\infty(0,T; L^2(\widetilde \Omega ))$ is dense in $L^2(0,T; L^2(\widetilde \Omega ))$. Now, I would like to know if $C^\infty(0,T; L^2(\Omega(t)))$ is dense in $L^2(0,T; L^2(\Omega(t)))$.
Here is my attempt: Let $v \in L^2(0,T; L^2(\Omega(t)))$ and $(v_n)_n \subset C^\infty(0,T; L^2(\widetilde \Omega ))$ such that $v_n \to v$ in $L^2(0,T; L^2(\widetilde \Omega ))$. This means that
$$\int_0^T\int_{\widetilde \Omega} |v - v_n|^2dy dt \to 0 \quad \text{for }n \to \infty.$$
This implies that $\|v(t, \cdot) - v_n(t, \cdot)\|_{L^2(\widetilde \Omega)} \to 0$ for every $t \in [0,T] \backslash E$, where $E$ has a zero measure. Therefore, there exists $N(t) \ge 0$ such that $\forall n \ge N(t)$, $\text{supp } v_n(t, \cdot) \subset \Omega(t)$. Now for $t \in E$, we can find a sequence $t_k \subset [0,T] \backslash E$ such that $t_k \to t$ and by continuity of the function $f$, for $k$ and $n$ sufficiently big, we have $\text{supp } v_n(t_k, \cdot) \subset \Omega(t) \cap \Omega(t_k)$ and as $t \mapsto v_n(t, \cdot)$ is continuous, $\text{supp } v_n(t, \cdot) \subset \Omega(t)$.
Now I would like to discard a finite number of $v_n$'s in my sequence so that $\text{supp } v(t, \cdot) \subset \Omega(t)$ for every $t \in [0,T]$. My problem is that $N(t)$ depends on $t$, such that for every $t$ I can discard a finite number for $n$'s in my sequence but how can I find a $N$ that does not depend on $t$ ? These $N(t)$'s could go to infinity for $t$ going to a certain value... Is it the right way to do that ? How can I conclude ?

EDIT: Actually the above argument is wrong. Indeed, when I deduce that $\forall n \ge N(t)$, $\text{supp } v_n(t, \cdot) \subset \Omega(t)$, it is as if I assumed that the support of $v(t, \cdot)$ was strictly include in $\Omega(t)$ while it is not true, we only have that the trace of $v$ is zero, but the support may not be strictly included. I should find a way to approach $v$ from inside $\Omega(t)$. I really have no idea how to do that..
 A: Take a function $v\in L^{2}(0,T;L^{2}(\tilde{\Omega}))$. Note that  $L^{2}(0,T;L^{2}(\tilde{\Omega}))$ can be identified with $ L^{2}((0,T)\times \tilde{\Omega})$. Hence we can find a Borel
representative $w:(0,T)\times\tilde{\Omega}\rightarrow\mathbb{R}^{2}$. Then by
Fubini's theorem and the change of variables $y=x-f(t)$,
\begin{align*}
\int_{0}^{T}\int_{\tilde{\Omega}}|w(t,x)|^{2}dxdt  & =\int_{0}^{T}\int%
_{\Omega+f(t)}|w(t,x)|^{2}dxdt\\
& =\int_{0}^{T}\int_{\Omega}|w(t,y+f(t))|^{2}dydt.
\end{align*}
Since the function $u(t,y)=w(t,y+f(t)))$ is measurable and in $L^{2}%
((0,T)\times\Omega)$, by standard density, given $\varepsilon>0$ we can find
$\varphi\in C_{c}^{\infty}((0,T)\times\Omega)$ such that
$$
\varepsilon>\int_{0}^{T}\int_{\Omega}|w(t,y+f(t))-\varphi(t,y)|^{2}%
dydt=\int_{0}^{T}\int_{\Omega+f(t)}|w(t,x)-\varphi(t,x-f(t))|^{2}dydt
$$
where we made the change of variables $y+f(t)=x$. By extending $\varphi$ to be zero outside $(0,T)\times\Omega$, we can assume that $\varphi\in C^\infty_c(\mathbb{R}^3)$.
Note that $\psi
(t,x)=\varphi(t,x-f(t))$ belongs to $C_c([0,T]\times \mathbb{R}^2)$ but it
is not $C^{\infty}$. Consider a sequence $f_{n}$ of functions in $C^{\infty
}([0,T])$ such that $f_{n}\rightarrow f$ uniformly in $[0,T]$. Then the
function $\psi_{k}(t,x)=\varphi(t,x-f_{k}(t))$ is in $C^{\infty}_c%
([0,T]\times \mathbb{R}^2)$. Moreover,
\begin{align*}
& \int_{0}^{T}\int_{\Omega+f(t)}|\varphi(t,x-f_{k}(t))-\varphi(t,x-f(t))|^{2}%
dydt\\
& =\int_{0}^{T}\int_{\Omega}|\varphi(t,y+f(t)-f_{k}(t))-\varphi(t,y)|^{2}dy
\end{align*}
again by the change of variables $x-f(t)=y$. Now $\varphi(t,y+f(t)-f_{k}%
(t))-\varphi(t,y)\rightarrow0$ uniformly in $[0,T]\times\overline{\Omega}$
as $k\rightarrow\infty$. Hence, for $k$ large
\begin{align*}
& \int_{0}^{T}\int_{\Omega+f(t)}|\varphi(t,x-f_{k}(t))-\varphi(t,x-f(t))|^{2}%
dydt\\
& =\int_{0}^{T}\int_{\Omega}|\varphi(t,y+f(t)-f_{k}(t))-\varphi(t,y)|^{2}%
dy<\varepsilon.
\end{align*}
In turn,
$$
\int_{0}^{T}\int_{\Omega+f(t)}|w(t,x)-\varphi(t,x-f_{k}(t))|^{2}%
dydt<4\varepsilon.
$$
