This is actually a question arising from statistics, regarding the proof of the Cramér–Rao inequality. The problem is only about measure-theory, though.
Let $T: (\mathcal{X}, \mathcal{F}) \rightarrow \bigl(\mathbb{R}, \mathcal{B}(\mathbb{R})\bigr)$ be a measurable function with $\mathcal{X} \subset \mathbb{R}$ a Borel set and $\mathcal{F}$ the Borel-sigma-algebra restricted on $\mathcal{X}$. Let $f: \mathcal{X}\times \Theta \rightarrow \mathbb{R}$ be also an $(\mathcal{X}, \mathcal{F})$-$\bigl(\mathbb{R}, \mathcal{B}(\mathbb{R})\bigr)$-measurable function for any fixed $\vartheta$ in an open real interval $\Theta$ and continuous for $\vartheta$.
In one step of the proof the textbook, with no explanation, claims that we can conclude from the set $$ N_\vartheta = \{x \in \mathcal{X} : T(x) \not= f(x,\vartheta)\}$$ having Lebesgue-measure zero for all $\vartheta$: $$\lambda(N_\vartheta)= 0 \quad \forall \vartheta \in \Theta$$ the following result: $$\lambda\bigl(\bigcup_{\vartheta \in \Theta}N_\vartheta\bigr)=0\,.$$
Question: If we want to fill in the intermediate steps, would the following be correct:
If $T(x) \not= f(x,\vartheta)$ for an $x\in\mathcal{X}$ and a $\vartheta\in\Theta$, define $$\varepsilon := |f(x,\vartheta)-T(x)|>0\,,$$ because $f(x,\vartheta)$ is continuous in $\vartheta$ there exists a $\delta >0$ so that $$|f(x,\vartheta)-f(x,\tilde{\vartheta})| < \varepsilon $$ for all $\tilde{\vartheta}$ with $|\vartheta - \tilde{\vartheta}| < \delta$. So choose a rational $\vartheta_0 \in ]\vartheta-\delta,\vartheta+\delta[ \cap\mathbb{Q}$.
For $\vartheta_0$ it must hold that $$|f(x,\vartheta)-f(x,\vartheta_0)| < \varepsilon = |f(x,\vartheta)-T(x)| \Rightarrow f(x,\vartheta_0) \neq T(x)\,.$$
So for any $\vartheta \in \Theta$ and any $x\in\mathcal{X}$ with $x\in N_\vartheta,$ we can find a $\vartheta_0\in\Theta\cap\mathbb{Q}$ with $x\in N_{\vartheta_0}$. This means that $$\bigcup_{\vartheta\in\Theta} N_\vartheta = \bigcup_{\vartheta \in \Theta \cap \mathbb{Q}} N_\vartheta \,.$$
Since $\lambda (N_\vartheta) = 0$ for all $\vartheta \in\Theta$, it follows with countable additivity of the Lebesgue-measure: $$\lambda\bigl(\bigcup_{\vartheta\in\Theta} N_\vartheta \bigr)= \lambda\bigl(\bigcup_{\vartheta \in \Theta \cap \mathbb{Q}} N_\vartheta\bigr) \leq \sum_{\vartheta\in\Theta\cap\mathbb{Q}} \lambda(N_\vartheta)= 0\,.$$
Could somebody please check that? Thanks.
