Can an area under the following function exist?

Question

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 Riemann 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 "integrable".

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 singularities which are neither continuous nor discontinuous.

(3) If removed the requirement 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 supremum 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 Darboux 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 Riemann integral so that $S:[a,b]\cap{A}\to\mathbb{R}$, where $A$ is dense subset of $\mathbb{R}$, instead of $S:[a,b]\to\mathbb{R}$?

Answer 1

In my definition, "the area under $S$" can exist in $[a,b]$ if the following requirements are met.

1. $S$ is a curve

2. 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.