Proving a variational inequality almost everywhere

Let $(\Omega_1,\mathcal{F}_1,\mathbf{P}_1)$ and $(\Omega_2,\mathcal{F}_2,\mathbf{P}_2)$ be two probability spaces (finite measure spaces). Suppose we have a random variable (measurable function) $X: \Omega_1\times \Omega_2\rightarrow \mathbb{R}$ such that $(\mathbf{P}_1⊗\mathbf{P}_2)(X<0)>0$.

(1) Can we conclude that there exists some measurable rectangle $A\times B$, with $\mathbf{P}_1(A)>0$ and $\mathbf{P}_2(B)>0$ such that $X<0$ on $A\times B$ ?

(2) Let $\Omega_2=[0,1] $ and $\mathbf{P}_2$ be the Lebesgue measure. Can we conclude that $$\underset{\mathbf{P}_2(B_\epsilon)=\epsilon}{\inf}\mathbf{E}^{\mathbf{P}_1}[\int_{B_\epsilon}X \mathrm{d}\mathbf{P}_2]\ge -o(\epsilon)$$is violated? In other words, is we have $$0\le \mathbf{E}^{\mathbf{P}_1}[\int_{B_\epsilon}X \mathrm{d}\mathbf{P}_2]+o(\epsilon)$$ for every choice of $B_\epsilon$, can we conclude that $X>0$, $\mathbf{P}_1⊗\mathbf{P}_2$ almost everywhere?

Remark: This question is derived from a crucial step in the proof of stochastic maximum principle, in which a variational inequality is claimed to hold $\mathbf{P}_1⊗\mathbf{P}_2$ almost everywhere.