Sufficient condition that the Infimum of an uncountable family of measurable functions is measurable

Question

Let $(\Omega, \mathcal{F})$ be a measurable space and $(f_r)_{r \in \mathbb{R}}$ be a family of $\mathcal{F}-\mathcal{B}(\mathbb{R})$-measurable functions $f_r : \Omega \to \mathbb{R}$. In general, the pointwise Infimum of an uncountable family is not measurable as seen here.

However, one of my homework assignments claims that if for every $\omega \in \Omega$ the function $\varphi_{\omega}: \mathbb{R} \to \mathbb{R}$ defined by $\varphi_{\omega}(r) := f_r(\omega)$ is continuous, then $\inf_{r \in \mathbb{R}}f_r$ is $\mathcal{F}-\mathcal{B}(\mathbb{R})$-measurable. I'm struggling to come up with a proof of this though, I can't really see how continuity of the $\varphi_{\omega}$ might come into play here. Any help would be greatly appreciated!

Answer 1

Spelling out geetha290krm's comment; we have (evaluating pointwise) $$ (\inf_{r\in\Bbb R}f_r)(\omega) = \inf_{r\in\Bbb R}(f_r(\omega)) = \inf_{r\in\Bbb R}(\varphi_\omega(r)) = \inf_{q\in\Bbb Q}(\varphi_\omega(q)) = \inf_{q\in\Bbb Q}(f_q(\omega)) = (\inf_{q\in\Bbb Q}f_q)(\omega), $$ so $$\inf_{r\in\Bbb R}f_r = \inf_{q\in\Bbb Q}f_q.$$

For passing from $\Bbb R$ to $\Bbb Q$, see Supremum over real interval equals supremum over rational 'interval' for continuos functions?. To finish from the above, see The Supremum and Infimum of a sequence of measurable functions is measurable.