Is it true that $\int_\Omega u(t,x)\;dx= \left(\int_\Omega u(s,x)\;dx\right)\bigg|_{s=t}$? Let $u \in L^2(0,T;L^2(\Omega))$. Define
$$f(t) = \int_\Omega u(t,x)\;dx$$ and $$g(t) = \left(\int_\Omega u(s,x)\;dx\right)\bigg|_{s=t}$$
for a.e. $t$.
Is it true that $f$ and $g$ are the same functions a.e. $t$? Can $u$ just be any function for this to hold, not just in some $L^p$ space (as long as integrals are sensible)?
Edit: I think it is trivially true. Maybe I'm just being stupid. 
 A: The function $u$ can be any function from the set in which $t$ takes value, into the space $\mathscr{L}^1(\Omega)$ of integrable functions on $\Omega$. Only for those $t$ where $u$ has a defined value $u(t,\cdot)$ in $\mathscr{L}^1(\Omega)$, 
the integral
\begin{equation*}
\int_\Omega u(t,x)dx
\end{equation*}
is defined. Considering this for every $t$ for which $u(t,\cdot)$ is defined, yields a function $f$ that maps $t$ to the integral
\begin{equation*}
f(t)=\int_\Omega u(t,x)dx.
\end{equation*}
For your function
\begin{equation*}
g(t)=\bigg(\int_\Omega u(s,x)dx\bigg)\bigg|_{s=t}
\end{equation*}
to make sense,
the integral
\begin{equation*}
\int_\Omega u(s,x)dx
\end{equation*}
must be defined,
that is, $s$ must be a value that is mapped by $u$ to the integrable function $u(s,\cdot)$ in $\mathscr{L}^1(\Omega)$,
and the integral of $u(s,\cdot)$ in terms of the above definitions is just $f(s)$.
Then writing
\begin{equation*}
g(t)=\bigg(\int_\Omega u(s,x)dx\bigg)\bigg|_{s=t}
\end{equation*}
is the same as writing
\begin{equation*}
g(t)=f(s)\big|_{s=t}.
\end{equation*}
So $f(t)$ and $g(t)$ are the same for every $t$ for which $u(t,\cdot)$ is defined.
