The negation of $(P)$ Let $(E,\mathcal{A},\mu)$ be finite measure space. Let $A\subset \mathbb{R}$ and the multifunction $F:E\to 2^{\mathbb{R}}$ 
We suppose that:
$$ 
 (P)~~:~~\exists N\in \mathcal{A} ~(\text{with }\mu(N)=0),\text{ such that: }\forall t\in E\smallsetminus N~:~F(t)\subset A
$$
Can we say that the negation of $(P)$ is :
$$
\exists N\in \mathcal{A} ~(\text{with }\mu(N)\neq 0) \text{ such that: } \forall t\in N~:~F(t)\nsubseteq  A
$$
If not, can you give me the negation of $(P)$. An idea please.
 A: I think it's good to think of the initial proposition in terms of $\mathcal B = \{N\in\mathcal A\,|\,\mu(N) = 0\}$.
Then the initial proposition is
$$
\exists N\in \mathcal{B}\text{ such that }\big( \forall t\in E\setminus N,~F(t)\subset  A\big).
$$
The negation is then simply
$$
\forall N\in \mathcal{B},\big( \exists t\in E\setminus N \text{  such that }F(t)\not\subset  A\big).
$$
A: The negation of a forall is an exists, and vice versa. So “not all humans are mortal” is logically the same as “there is a human who is not mortal” and “there is no god who is mortal” is logically the same as “all gods are not mortal.”
More formally, $$\neg(\exists x \in Y: f(x))~~\Leftrightarrow~~\forall x \in Y: \neg f(x),\\
\neg(\forall x \in Y: f(x))~~\Leftrightarrow~~\exists x \in Y: \neg f(x).
$$
In your particular case you have 
$$
\neg(\exists N \in \mathcal A: \mu(N)=0~\wedge~\forall t\in N: F(t)\subset A)
$$
Distributing this gives
$$
\forall N \in \mathcal A: \mu(N)\ne 0~\vee~\exists t\in N: F(t)\not\subset A
$$
We could then rewrite that last part as $\mu(N) = 0 ~\rightarrow~ \exists t\in N: F(t)\not\subset A$ and then incorporate the “with $\mu(N) = 0$” into the “forall” clause, if you wanted to instead do like the other answer did and view  the original (P) as restricting focus to a set $\mathcal B = \mathcal A \cap \operatorname{ker} \mu$ : you can indeed see this as saying for all elements $N$ of $\mathcal B$ there is some $t$ in $N$ such that $F(t)$ is not a subset of $A$. In this way a domain restriction on an “exists”, manifesting as a conjunction, becomes (on negation) an implication, which is a domain restriction on a “forall.”
