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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?

Alphie
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    The statement is "for all $\delta > 0$ the set ${\omega: ; X(\omega) \geq \delta }$ has measure zero". So the negation is "there exists $\delta > 0$ such that the set ${\omega: ; X(\omega) \geq 0 }$ does NOT have measure zero". Past this, I can't help much because I don't know enough about the generalized notions you're dealing with (e.g. is every subset of a measure zero set a measure zero set). Possibly of interest to you (and probably not!), see answer 1 and answer 2. – Dave L. Renfro Feb 12 '21 at 19:25

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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 $\mathbb P(A)>0$.

Will
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