Let $1\leq p\leq \infty$, $I\subseteq\mathbb R$ an open interval and $f\in L^p(I)$ and $\Phi\colon \mathbb R\to\{\text{true},\text{false}\}$ be a predicate.
Sometimes it is written something like that: $$ \Phi(f(x))\qquad\text{almost everywhere in $I$}.\tag{P} $$
I think this means the following: $$ \exists \tilde f\in f\colon\quad \lambda(\{ x\mid \neg \Phi(\tilde f(x)\}) = 0 \tag{P1}$$, where $\tilde f$ is a representant of $f$ and $\lambda$ is the Lebesgue measure. This means its negation would be $$ \forall \tilde f\in f\colon\quad \lambda(\{ x\mid \neg \Phi(\tilde f(x)\}) \neq 0 \tag{N1}$$
Another way of putting (P) would be $$ \exists \tilde f\in f\colon\quad \exists M\in N(I)\colon\quad\forall x\in I\setminus M\colon\quad \Phi(\tilde f(x))\tag{P2}$$ where $N(I)$ is the set of all Lebesgue measureable subsets of $N$ with vanishing Lebesgue measure. Then its negation would be $$ \forall \tilde f\in f\colon\quad \forall M\in N(I)\colon\quad\exists x\in I\setminus M\colon\quad \neg \Phi(\tilde f(x))\tag{N2}$$
My question are:
- Are (P1) and (P2) correct interpretations of (P)?
- If so, are they equal?
- Are (N1) and (N2) correct negations of (P1) and (P2)? (I'm pretty sure about that)
- If so, how can they be interpreted
- Last but not least: How does the negation of (P) correspond to $$\neg \Phi(f(x))\qquad\text{almost everywhere in $J\subseteq I$, where } \lambda(J)\neq 0. \tag{N}$$
Firstly, your $p$ is irrelevant. We may as well just be talking about measurable functions with properties that are true a.e.
I would say that equivalence of (P1) and (P2) is pretty straightforward. Assume one, and try to prove the other, and vice versa. However, I would go further and say that in fact, both of your (P1) and (P2) are a bit weaker than the truth, in that if they hold for any representant, they hold for all. This is because any two representants differ only on a set of measure $0$. But regardless, they are equivalent.
I think (N1) is a little wrong. In particular there are no guarantees that the set where the proposition is false is measurable. You could say that for all representants $\tilde f$ and $A\in\sigma(\mathbb{R})$ with $\{x | \neg\Phi(\tilde f(x))\} \subseteq A$, $\lambda(A) > 0$. This is much more obviously equivalent to your second condition.
(N) is a stronger condition than not (P). Its saying that the set where the proposition fails contains a non null set, whereas the correct interpretation of not (P) is that the set where the proposition fails cannot be contained in a null set.
Hope there's some clarity in that for you.