Suppose I have two functions $f, g \in L^2(0,T;L^2(\Omega))$ where we have some bounded domain $\Omega$.
Suppose that $$\text{for almost all $t$,}\quad f(t) \leq g(t) \quad\text{almost everywhere in $\Omega$}.$$
Is this the same as saying $$f(z) \leq g(z) \quad\text{almost every $z \in [0,T]\times \Omega$}?$$
I think it is true when $f$ and $g$ are both non-negative functions, because then I can probably use Fubini theorems. But in the general case, I don't know.
The non-negativity of $f$ and $g$ has nothing to do with this. It's all about the set $$N = \{(t,x)\in [0,T]\times\Omega : f(t,x)>g(t,x) \}$$ The first statement is that almost every time slice of $N$ has zero measure. The second statement is that $N$ has zero product measure. These are equivalent by the Fubini-Tonelli theorem applied to $\chi_N$.