Measurability of function related to a Riemannian metric Let $(\mathcal{M},g)$ denote a Riemannian manifold. Let $T \subset \mathbb{R}$ be a compact subset, and let $\mu:T \to \mathcal{M}$ be a measurable function, where $\mathcal{M}$ is equipped with its Borel $\sigma$-algebra. Let $V,U$ be measurable vector fields along $\mu$, ie. $V:T \to TM$ where $V(t) \in T_{\mu(t)}M$ for all $t \in T$. I wish to show that the function $f:T \to \mathbb{R}$ given by
$$
f(t) = \langle V(t),U(t) \rangle_{\mu(t)} = g_{\mu(t)}(V(t),U(t))
$$
is measurable.
I know that if $V,U$ are smooth vector fields on $\mathcal{M}$ then $p \mapsto \langle V,U \rangle_{p}$ is a smooth map from $\mathcal{M}$ to $\mathbb{R}$. I suspect it follows by noting that
$$
f = T \xrightarrow{\mu} \mathcal{M} \xrightarrow{g} \mathbb{R}  $$
where $g(p) =\langle V_p,U_p \rangle_p$ but I am not 100% confident, since we do not have any guarantee that $g$ would be measurable.
 A: I was able to answer the question when assuming $V:T \to TM$ and $U:T \to TM$ are extendible vector fields, i.e. under the assumption that there exists measurable vector fields $\tilde{V}:\mathcal{M} \to TM$ and $\tilde{U}:\mathcal{M} \to TM$ st. for all $t \in T$, $V(t) = \tilde{V}(\mu(t))$.
This is shown by noting that the function
$$
f(t) = \langle V(t),U(t) \rangle_{\mu(t)} = \langle \tilde{V}(\mu(t)),\tilde{U}(\mu(t)) \rangle_{\mu(t)} = \langle \tilde{V}, \tilde{U} \rangle \circ \mu (t)
$$
is measurable if and only if  $\langle \tilde{V}, \tilde{U} \rangle: \mathcal{M} \to \mathbb{R}$ is measurable, since $\mu$ is measurable by assumption. Note that the function $\langle \tilde{V}, \tilde{U} \rangle :\mathcal{M} \to \mathbb{R}$ is given by $\langle \tilde{V}, \tilde{U} \rangle (p) = \langle \tilde{V}(p), \tilde{U}(p) \rangle_p $. To show measurability of $\langle \tilde{V}, \tilde{U} \rangle$ it is enough to find a countable open cover $\mathcal{M} = \cup_{n \geq 1} U_n$ and show that $\langle \tilde{V}, \tilde{U} \rangle |_{U_n}:U_n \to \mathbb{R}$ is measurable for each $n$. Since if  that holds, then we have that for any $a \in \mathbb{R}$,
$$
\langle \tilde{V}, \tilde{U} \rangle^{-1}((a,\infty))=
\{p \in \mathcal{M} : \langle \tilde{V}, \tilde{U} \rangle (p) \in (a, \infty) \} 
=
\cup_{n \geq 1} \{p \in U_n : \langle \tilde{V}, \tilde{U} \rangle |_{U_n}(p) \in (a, \infty)
\}
$$
Thus, if each $\langle \tilde{V}, \tilde{U} \rangle |_{U_n}$ is measurable, then all sets in the union are measurable.
Now, to find such open cover, fix any atlas $\{U_{\alpha},\phi_{\alpha}\}$ over $\mathcal{M}$ and extract from this a countable cover. This is possible since $\mathcal{M}$ is second countable and therefore Lindelof. Thus, if we can show $\langle \tilde{V}, \tilde{U} \rangle |_{U_n}$ is measurable we are done.
Now, we have that the coordinate-form of $\langle \tilde{V}, \tilde{U} \rangle : U_n \to \mathbb{R}$ is given by
$$
\langle \tilde{V}, \tilde{U} \rangle(x) =\sum g_{i_1,i_2}(x)dx^{i_1}\otimes dx^{i_2}(\tilde{V}(x),\tilde{U}(x)) \\
= \sum g_{i_1,i_2}(x)\tilde{V}^{i_1}(x)\tilde{U}^{i_2}(x)
$$
which shows that the coordinate-version of $\langle \tilde{V}, \tilde{U} \rangle$ is measurable, since all the involved component functions are measurable. This, in turn, implies measurability of the original $\langle \tilde{V}, \tilde{U} \rangle : U_n \to \mathbb{R}$ since $\langle \tilde{V}, \tilde{U} \rangle \circ \phi_n^{-1}$ is measurable iff $\langle \tilde{V}, \tilde{U} \rangle$ is measurable, since $\phi_n$ is a homeomorphism and thus guaranteed measurable.
