Let $F:A\to \mathbb{R}$ be a function where $A = I_1\times \dots\times I_n\subseteq \mathbb{R}^n$ is a not necessarily open rectangle. Suppose that partial derivatives $F_i = \frac{\partial F}{\partial t_i}:A\to\mathbb{R}$ exist for all $i = 1, \dots, n$ in the sense that we take one-sided derivatives on the edges of the intervals. Further suppose that $F_i$ are continuous, so that we can say that "$F$ is $C^1$ on $A$".
If now $f_i:J\to I_i$ are differentiable where $J\subseteq \mathbb{R}$ is an interval, again in the sense of one-sided differentiability at the end-points, does the chain rule for the function $G(t) = F(f_1(t), ..., f_n(t))$, $G:J\to\mathbb{R}$ hold true?
That is, does $G'(t) = \sum_{i=1}^n F_i(f_1(t), ..., f_n(t))\cdot f_i'(t)$ for $t\in J$?
Context:
I'm trying to prove Leibniz rule for gauge integrals and I'm trying to use above theorem, but I need it to work for closed rectangles so that I can deduce differentiability at the edges of the interval as well. I never saw anything similar to this, and it makes sense since differentiability is defined on open sets, so reasoning of the form "$F$ is $C^1$ on $A$, so differentiable on $A$, so chain rule holds" is not available.
We can prove this using a standard argument. For $t\in J, \Delta t\neq 0$ we have $$G(t+\Delta t)-G(t) = \sum_{i=1}^n \left(F(\dots, f_i(t+\Delta t), \dots) - F(\dots, f_i(t), \dots)\right)$$ with the understanding that terms left to $i$th coordinate have argument $t+\Delta t$, and to the right of $i$th coordinare have argument $t$. Here $\Delta t > 0$ if $t$ is a left endpoint and $\Delta t < 0$ if $t$ is a right endpoint.
Then from mean value theorem applied to $x_i\mapsto F(\dots, x_i, \dots)$ $$F(\dots, f_i(t+\Delta t), \dots) - F(\dots, f_i(t),\dots) = (f_i(t+\Delta t)-f_i(t))\cdot \partial_i F(\dots, s_i, \dots)$$ where $s_i$ lies between $f_i(t+\Delta t)$ and $f_i(t)$. Dividing both sides by $\Delta t$ and letting $\Delta t\to 0$ we obtain $f_i'(t)\cdot \partial_i F(f_1(t), ..., f_n(t)) $ from continuity of $\partial_i F$ and $f_1, ..., f_n$. Thus $$G'(t) = \sum_{i=1}^n f_i'(t)\cdot \partial_i F(f_1(t), ..., f_n(t)),\ t\in J.$$