Given a function $u \in L^{1}(\mathbb{R} \times (0, \infty))$ does there exist a function $v \in C((0,\infty); L^1(\mathbb{R}))$ such that $v=u$ a.e. in $\mathbb{R} \times (0, \infty).$
If not is there any additional assumptions on $u$ which ensures existence of such a $v$?
The answer is NO. Take $u(x,t)=e^{-|x|}g(t)$ where $g$ is a positive integrable function on $(0,\infty)$. If $v$ exist as stated then there is at least on $x$ such that $e^{-|x|}g(t)=v(x,t)$ for almost all $t$. But then $t \to \|v(x,t)\|_1$ is continuous which implies $t \to g(t)$ is almost everywhere equal to a continuous function. But this is not true in general. An example of a function $g$ for which this fails is $I_C$ where $C$ is a fat Cantor set. Trying to get a continuous function which is almost everywhere equal to a given integrable function seems pretty unrealisitic and I cannot think of any simple sufficient conditions for this.