Let $(S,\mathcal{A},μ)$ be an underlying probability space , $X:(S,\mathcal{A})\to (\mathcal{X},\mathcal{B}(\mathcal{X}))$ . Let $ (\mathcal{X},\mathcal{B}(\mathcal{X})) = (\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n))$ , $X=(X_1,X_2,\dots,X_n)$ . Let $X_i$ be i.i.d. continuous Uniform($0,θ$) . We do not know θ but we know that $θ\in Ω = ]0,\infty[$ , hence $ μ \circ X^{-1} \in \lbrace P_θ | θ\in Ω \rbrace$ . Let $T=X_{(n)}$ be the maximum value of the elements of $X$ . Let $g:(\Omega_0 , \mathcal{A}_0)\to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ . To prove that T is complete we have to prove that the following statement is TRUE : $$ \forall g \bigg\lbrack \big(\forall θ\left(\operatorname{E}_{P_θ}[g(T(x))] = 0\right)\big) \rightarrow \big(\forall θ\left(P_θ (\lbrace g(T(x))=0 \rbrace ) = 1 \right)\big) \bigg\rbrack$$ Here it has been proved that : $$\forall θ ~λ(\lbrace t\in [0,θ] | \neg(g(t) = 0) \rbrace) = 0$$ What I fail to understand is how that implies $$ \forall θ ~\left(P_θ (\lbrace g(T(x))=0 \rbrace ) = 1 \right) $$ .
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Let $\lbrace t\in [0,θ] | \neg(g(t) = 0) \rbrace = \lbrace t_i | i\in I \rbrace , \#I\le \aleph_0$ . It is valid that : $ \forall i ~P_θ(\lbrace T(x) = t_i \rbrace) = 0 $ and that is because $F_T (t)$ is an absolutely continuous c.d.f. . Hence $$P_θ(\mathcal{X}\setminus \lbrace x\in \mathcal{X}|\exists i \in I ~T(x)= t_i \rbrace) = 1 \implies g(T(x)) = 0 ~~P_θ - a.s. $$
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