Why is $$u: (G := [0,r] \times \mathbb{R}^n) \to \mathbb{R}^n: (x,z) \mapsto \int_0^x f(v(s,z))ds$$ continuous, where we endow the domain with the product metric, and where
$$v: G \to \mathbb{R}^n; f: \mathbb{R}^n \to \mathbb{R}^n$$ are continuous functions?
Attempt: Probably this follows easily from some fact about the product metric/topology, but I can't see it.
I was messing around with the fundamental theorem of calculus, and it is easy to see that for every fixed $z$ the associated function $u_z$ is differentiable, and therefore continuous, but I don't think this is sufficient to deduce continuity of $u$.
Context of this problem: book on differential equations
Let $\varepsilon>0$. First notice that, since $v:G\to \mathbb{R}^n$ and $f:\mathbb{R}^n\to\mathbb{R}^n$ are continuous, so is $h \equiv f\circ v : G\to \mathbb{R}^n$. Let $e_x(\delta),e_z(\delta)$ any functions such that $e_z(\delta),e_x(\delta)\to 0$ as $\delta\to 0$. Continuity for $u$ is equivalento to say that $$ u(x+e_x(\delta),z+e_z(\delta))\to u(x,z) \quad \text{ as } \quad \delta\to 0.$$ Now we can write $$ u(x+e_x(\delta),z+e_z(\delta)) = \int_0^{x+e_x(\delta)}h(s,z+e_z(\delta))ds =\\ \int_0^{x}h(s,z+e_z(\delta))ds + \int_x^{x+e_x(\delta)}h(s,z+e_z(\delta))ds = A(\delta)+B(\delta). $$ Since $h$ is continuous over $[0,r]\times B_1(z)$ (I'm looking for a set that would contains points $(s,z+e_z(\delta))$ for $\delta$ small enough), then we can apply Dominated Convergence Theorem and deduce that $$ A(\delta)\xrightarrow{\delta\to 0} \int_0^x h(s,x)ds. $$ On the other side, $h$ is bounded over compact sets because it is continuous, so we find that $$ |B(\delta)| \leq \int_x^{x+e_x(\delta)}|h(s,z+e_z(\delta))|ds \leq \|h(s,x)\|_{L^\infty([0,r]\times B_1(z))} e_x(\delta) \xrightarrow{\delta\to 0}0. $$ This shows that $u(x+e_x(\delta),z+e_z(\delta))\to u(x,z)$, i.e. $u$ is continuous.