Let $T>0$. We take a look at the function $f:[0, T] \rightarrow \mathbb{R}$, $$ f(t) := \min_{x \in I(t)} o(x) $$ where $o:\mathbb{R} \rightarrow \mathbb{R}$ is positive, continuous, bounded (from below and above) and $\displaystyle \min_{x \in \mathbb{R}} o(x)$ exists. If $I(t) = [g(t), h(t)] \subseteq \mathbb{R}$ where $g:[0, T] \rightarrow [-\infty, \infty)$ and $h:[0, T] \rightarrow (-\infty, \infty]$ are measurable functions, is $f$ measurable as well?
What comes to mind is that we can find at least some minimizer for $o$ on $I(t)$. Whether this minimizer is measurable in $t$ is not clear to me. Also, I know that the $\inf$ of a sequence of functions is measurable, but I do not expect this to help. This is a very delicate case, so I have not found it in the literature yet. I would also be happy if someone had a reference. Thank you.
Notice that $f(t) = F(g(t), h(t))$, where $$ F(a, b) = \min_{a \leq x \leq b} o(x). $$
To see that $F$ is measurable, we have that, since $o$ is continuous, $$ \{(a, b) : F(a , b) \geq L\} = \bigcap_{n=1}^\infty \bigcup_{\substack{a, b \in \mathbf{Q} \\ \forall x\in [a, b], o(x) \geq L}} (a - n^{-1}, a + n^{-1}) \times (b - n^{-1}, b + n^{-1}). $$
This, $f$ is a composition of measurable functions and so is itself measurable.