Is there a way to integrate over reals so that the uncountable sum is finite?

Question

I've been doing some research on this and I saw that a bounded integration over reals evaluates to infinity. This makes total sense, such as in the case:

$$ y = x $$

If we take the typical integral with bounds [0, 1], we obtain:

$$ \int_{0}^{1} x dx = \frac{1}{2} x^2 |_{0}^{1} = \frac{1}{2} - 0 = \frac{1}{2} $$

However, we are taking dx to mean an infinitesimal amount. If we allowed dx = 1, then we would be summing over reals [0, 1] and the value would be infinity.

I can think of cases where this value would be finite. Suppose now we have:

$$ y = 0 $$

where there are discontinuities at certain values for x. In other words, there are random values of x (not all reals) where:

$$ y \neq 0 $$

For example:

$$ y = \left\{ \begin{array}{ll} 1 & x = \frac{1}{3} \\ 2 & x = \frac{1}{2} \\ 3 & x = \frac{\sqrt{2}}{2} \\ \end{array} \right. $$

Thus, the sum would be finite (S = 6) and we could sum over reals. Is there a way to do this like integration (sum up all values of y)? Any help would be appreciated.

Update:

I have been doing more research on this topic and I actually figured out that a good way to do this is substitute the floor of x for x in the function you are trying to integrate. But this brings up another question: what is the integral of the floor of x? Furthermore, what is the integral of a power of the floor of x? I saw in one of the answers here that:

$$ \int \;\lfloor x \rfloor dx = x\lfloor x \rfloor - \frac12 \lfloor x \rfloor(\lfloor x \rfloor + 1)\; $$

But the person did not provide a proof. I do not quite understand how they arrived at that answer. I tried Googling it, but I can't seem to find a proof. Can anyone provide one please? Additionally, can you provide how you would take the integral of a power of the floor of x, such as:

$$ \int \;\lfloor x \rfloor^n dx $$

where n is a natural number. I'd appreciate it.

Answer 1

In theory of Lebesgue integration there is no such concern, in fact, you usually have the following definition: $$\int_A f(x)\mathrm{d}x := \int 1_A(x)f(x)\mathrm{d}x$$ where $1_A(x) = \begin{cases} 1 & x\in A \\ 0 & x\notin A \end{cases}$ and right hand side is well understood as an integral on the whole reals.

Answer 2

There are contexts in which it is acceptable to sum over an uncountable set like $R$. You just have to guarantee that the number of nonzero addends is countable for the sum to make sense. For instance, if $P$ is a degree-$n$ polynomial, then $$\sum_{x\in R} \text{mult}_P(x) \leq n,$$ where $\text{mult}_P(x)$ is the multiplicity of $x$ as a root of $P$. (In all but finitely many cases, the multiplicity is zero.) But this isn't an integral, and nothing like it ever will be. Thinking of an integral as a sum over uncountably many infinitesimal values is helpful sometimes but ultimately incorrect. There are various ways of defining integration, but commonly the integral can be thought of as a limit of increasingly granular countable sums. For example, the Reimann integral is the limit of the Reimann sums as the rectangle base width goes to 0. You will only get so far if you try to extend the properties of finite or countable addition to this type of operation.