Let $u \in L^2(\Omega \times (0,T))$ where $\Omega \subset \mathbb{R}^n$ is an open bounded set and $T>0$.
Consider the operator $$F(u)=\begin{cases} u \quad \text{ if } \vert u(x,t) \vert \leq k \text{ for almost every } (x,t)\in \Omega \times (0,T)\\ k \quad \text{ if } u(x,t) > k \text{ for almost every } (x,t)\in \Omega \times (0,T)\\ -k \quad \text{ if } u(x,t)<-k \text{ for almost every } (x,t)\in \Omega \times (0,T) \end{cases}$$
Is $F$ well-defined for $u \in L^2(\Omega \times (0,T))$?
Here is my approach:
First, I had doubts that the regularity $L^2(\Omega \times (0,T))$ is not enough, since we cannot talk about pointwise values of $u$ in this case. But then again, I have tried to use the fact that since $u \in L^2(\Omega \times (0,T))$ then it is finite almost everywhere in $\Omega \times (0,T)$ and therefore it must satisfy one of the three cases indicated in the definition of $F$ and hence $F(u)$ is well-defined.
No, I do not think it is. And if by your observation that $u$ being square integrable implies it being finite almost everywhere you mean that it must be bounded (i.e., $|u(x,t)|\leq k$ for some $k$) almost everywhere, then that is not correct.
For example, consider the case $\mathbb{R}^n = \mathbb{R}$ and $T = 1$ and choose $\Omega$ sufficiently large (any bounded open $\Omega$ containing $[0,\pi^4/90]$ will do). Consider the sequence $(a_n)^\infty_{n=0}$ defined inductively by \begin{equation*} \begin{aligned} a_0 &= 0, \\ a_n &= a_{n-1} + n^{-4}\quad\text{for }n > 0, \end{aligned} \end{equation*} and let \begin{equation*} u(x,t) = \sum^\infty_{n=1}n 1_{[a_{n-1},a_{n})}(x). \end{equation*} Then (using some particular values of the Riemann zeta function) we find \begin{equation*} \begin{aligned} a_0 &= 0 \\ a_1 &= 0 + 1^{-4} \\ a_2 &= 0 + 1^{-4} + 2^{-4} \\ a_3 &= 0 + 1^{-4} + 2^{-4} + 3^{-4} \\ &\vdots \\ \lim_{n\to\infty}a_n &= \sum^\infty_{n=1}n^{-4} = \frac{\pi^4}{90}, \end{aligned} \end{equation*} hence $(a_n)^\infty_{n=0}$ lies in $\Omega$, and \begin{equation*} \int_{\Omega\times(0,T)}|u(x,t)|^2 dxdt = \sum^\infty_{n=1}n^2(a_{n+1}-a_n) = \sum^\infty_{n=1}n^2n^{-4} = \sum^\infty_{n=1}n^{-2} = \frac{\pi^2}{6}. \end{equation*} So $u$ is well-defined on $\Omega\times(0,T)$ and square integrable, but for each positive integer $k$ we have that neither $|u(x,t)|\leq k$ almost everywhere nor $u(x,t) > k$ almost everywhere, and $u(x,t) < -k$ for no $(x,t)\in\Omega\times(0,T)$. This leaves $F(u)$ undefined.