Can the Gauge Integral exist on a function defined on a countable set? What would it equal?
I was wondering because if we took the Darboux Integral of $f:C\cap[a,b]\to\mathbb{R}$ where $C$ is a countable dense set and $f$ is continuous in $C$, the integral is the same as the Darboux Integral of $f:[a,b]\to\mathbb{R}$.
The definition of all known integrals state $f$ has to be defined in $[a,b]$ ? Why so? What are they trying to avoid?
I will assume that in your context, $f$ is a function defined (and continuous, for simplicity) on all of $[a,b]$. If this is not a fair assumption let me know and I will edit (or delete) my answer.
Let $C$ be a countable dense subset of $\mathbb{R}$. As you noted, applying the definition of the Darboux integral (in particular with Darboux sums) to $f$ while only considering the domain $C \cap [a,b]$ yields the same result as the Darboux integral of $f$. For lack of better notation, I will denote the limiting Darboux sum of $f$ when restricted to $C \cap [a,b]$ by:
\begin{align*} \int_{C \cap [a,b]} \left. f \right|_{C \cap [a,b]} \end{align*}
So, what you have noticed is that:
\begin{align*} \int_{C \cap [a,b]} \left. f \right|_{C \cap [a,b]} = \int_{[a,b]} f \end{align*}
(I have not checked that this is always true, but I suspect that it is if $f$ is continuous on $[a,b]$.)
The reason, then, that we don't use this as a definition for the integral of $f$ over $C \cap [a,b]$ is that we lose a nice property of integration: additivity of the domain.
Let $g_1 : [a,b] \rightarrow \mathbb{R}$ and $g_2 : [c,d] \rightarrow \mathbb{R}$ be such that $[a,b] \cap [c,d] = \varnothing$. Define $g : [a,b] \cup [c,d] \rightarrow \mathbb{R}$ as follows: \begin{align*} g(x) = \left\{ \begin{matrix} g_1(x) & \text{if } x \in [a,b] \\ g_2(x) & \text{if } x \in [c,d] \end{matrix} \right. \end{align*} Then we have the following nice property:
\begin{align*} \int_{[a,b]\cup[c,d]} g = \int_{[a,b]} g_1 + \int_{[c,d]} g_2 = \int_{[a,b]} \left. g \right|_{[a,b]} + \int_{[c,d]} \left. g \right|_{[c,d]} \end{align*}
As it turns out, with your definition, this no longer holds. Consider the following example. Let $C_1 = \mathbb{Q}$, and let $C_2 = \mathbb{Q} + \sqrt{2} = \{ q + \sqrt{2} : q \in \mathbb{Q}\}$. Then both $C_1$ and $C_2$ are countable dense sets, and their intersection is $\varnothing$. Therefore their union, $C = C_1 \cup C_2$, is also countable and dense.
But then by your definition, we should have:
\begin{align*} \int_{C_1 \cap [a,b]} \left. f \right|_{C_1 \cap [a,b]} = \int_{C_2 \cap [a,b]} \left. f \right|_{C_2 \cap [a,b]} = \int_{C \cap [a,b]} \left. f \right|_{C \cap [a,b]} = \int_{[a,b]} f \end{align*}
And so, as long as $\int_{[a,b]} f \neq 0$, we end up with:
\begin{align*} \int_{C_1 \cap [a,b]} \left. f \right|_{C_1 \cap [a,b]} + \int_{C_2 \cap [a,b]} \left. f \right|_{C_2 \cap [a,b]} \neq \int_{C \cap [a,b]} \left. f \right|_{C \cap [a,b]} \end{align*}
Since additivity of the domain is one of the most important features of the integral, we choose not to use the definition you proposed. Instead, we agree that the integral over a countable dense set is either undefined (as per the Darboux integral) or zero (as per the Lebesgue integral).