The paper am reading proves a statement of the following form:
For all $\delta>0$, $$ X<\delta \hspace{1cm} \text{almost surely}$$
where $X$ is real-valued function on some probability space $(\Omega,\mathcal{F},P)$. The proof is by contradiction. It begins by saying:
Suppose there exist a set $A\in \mathcal{F}$ with $P(A) > 0$, and a constant $\delta>0$, such that $$ X(ω) ≥ δ $$ for any $\omega \in A$.
The proof then proceeds by showing that this assumption leads to a contradiction. But I don't see why the second statement is the negation of the original one.
Any ideas?
Suppose that the set $\{X<\delta\}$ is measurable. Then $X<\delta$ almost surely means that $\mathbb P(\{X<\delta\})=1$, or equivalently $\mathbb P(\{X\ge\delta\})=0$. Its negation is therefore $\mathbb P(\{X\ge\delta\})>0$, or equivalently there exists a measurable set $A\subset\{X\ge\delta\}$ such that $\mathbb P(A)>0$.
Isn't $X$ a random variable? In that case, $\{X<\delta\}$ is measurable. If $\{X<\delta\}$ is not measurable, then your proof is not correct.
Take for example $(\Omega,\mathcal F)=([0,1],\{\emptyset,[0,1]\})$ and $\mathbb P(\emptyset)=0$, $\mathbb P([0,1])=1$. Let $X:x\in[0,1]\mapsto x$ and $\delta=\frac12$. Then $\{X<\delta\}=[0,\frac12)$, so we do not have that $X<\delta$ almost surely. We do not have either the existence of a measurable set $A\subset\{X\ge\delta\}=[\frac12,1]$ such that $\mathbb P(A)>0$.