Can an area under the following function exist? 
Consider the function $S: [a,b] \cap \Bbb Q \to \Bbb R$ defined by $S(x)= x^2$. Does it make sense to speak about the area under the graph of $S$?

The Riemman integral and Lebesgue integral exists only if $S:[a,b]\cap \mathbb{R}\to\mathbb{R}$. However $S(x)$ has many properties could make it "integratable".
Properties of $S(x)$:
(1) The function is continuous.
The limit of $S(x)$ exists at any $c\in[a,b]$, if every $c$ is a cluster point of $\mathbb{Q}$. Since $\mathbb{Q}$ is a dense in $\mathbb{R}$ the previous requirement is met and the limit exists at any point in the interval.
The continuity at $c$ would then exist if
\begin{equation}
\underset{x\in \mathbb{Q}}{\underset{x \to c}\lim} S(x)= S(c)
\end{equation}
which is true for $c\in \mathbb{Q}$. Since $S(x)$ is continuous in its domain, $S(x)$ is a continuous function.
(2)The function has no discontinuities
Continuous or discontinuous points exist in the domain. As mentioned earlier, the domain of $S(x)$ is continuous and hence there are no discontinuities.
Furthermore, the points where $x\not\in 
\mathbb{Q}$ are removable singluarities which are neither continuous nor discontinuous.
(3) If removed the requirment that $S:[a,b]\to{\mathbb{R}}$, for the Darboux sums to exist; the upper and lower sum would converge.
The function $S(x)$ is continuous on a set dense in $\mathbb{R}$. Hence if we divided $[a,b]$ into sub-intervals of partitions, no matter how small the sub-intervals, the suprenum and infimum of $S$ at every sub-interval would exist.
\begin{equation}
\sum_{j=1}^{n-1}\inf_{[x_j,x_{j+1}]}S(x)<\text{Area of $S(x)$}<\sum_{j=1}^{n-1}\sup_{[x_j,x_{j+1}]}S(x)
\end{equation}
Which is the same as
\begin{equation}
\sum_{j=1}^{n-1}\inf_{[x_j,x_{j+1}]}x^2<\text{Area of $S(x)$}<\sum_{j=1}^{n-1}\sup_{[x_j,x_{j+1}]}x^2
\end{equation}
If the darbox sum did not require $S:[a,b]\to\mathbb{R}$ then $\text{Area of $S(x)$}= \int_{a}^{b} x^2 dx$.
Conclusion
Could an area exist, without extending the domain of $S$ to the entire interval?
Should we change the definition of the Riemman integral so that $S:[a,b]\cap{A}\to\mathbb{R}$, where $A$ is dense subset of $\mathbb{R}$, instead of $S:[a,b]\to\mathbb{R}$?
 A: In my definition, "the area under $S$" can exist in $[a,b]$ if the following requirements are met. 

  
*
  
*$S$ is a curve
  
*The sets defined by $S$ have measure of one compared to sets outside of $S$.

The definition of a curve is the following.

If $S:[a,b]\cap{A_n}\to\mathbb{R}$, such that $S$ is continuous in the interval and $A_n$ is a dense subset of $\mathbb{R}$, then $S$ is a curve.

Note that a functions may have more than one curve. Consider Dirichlet's function.
$$f(x)=\begin{cases}1 & x\in\mathbb{Q}\\ 0 & x\not\in{\mathbb{Q}} \end{cases}$$
We know $y=1$ is continuous everywhere and ${\mathbb{Q}}$ is a dense subset of $\mathbb{R}$ and $y=0$ is continuous everywhere and irrationals is a dense subset of $\mathbb{R}$. Hence two curves exist.
The second requirement involves the measure of the sets defined by $S$. If $A_n$ has a measure of one compared to defined points outside the curve, then an area under $S$ has to exist.
In the case of the Dirichlet function, since the irrationals have a Lebesgue measure of $1$, the area under $0$ exists. 
From this, I make my case that a function made of several curves does not have to be defined everywhere to have an area under one curve. Due to this, I am creating new measures where the area under curves of disconnected sets can exist.
