Wikipedia:Reference desk/Archives/Mathematics/2017 July 19

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July 19

Integrating over countable set

As a note, this is not an assigned problem, just something I was thinking about. Suppose $f(x)$ is strictly positive and continuous over an open interval $S=(a,b)$, so that $\int _{S}f(x)dx>0$. We commonly encounter integration over these uncountably infinite sets like $S$ in evaluating integrals. Note that if we create the finite set $D=\{x_{k}:x_{k}\in S,k=1,...,n\}$, for example, some $n$ integers lying in $(a,b)$, it appears that $\int _{D}f(x)dx=\sum _{k=1}^{n}\int _{x_{k}}^{x_{k}}f(x)dx=0$. So my question is, is it ever possible to have a finite, or countably infinite, subset $S^{*}\subset S$, such that $\int _{S^{*}}f(x)dx>0$? 50.101.93.214 (talk) 00:09, 19 July 2017 (UTC)

@50.101.93.214: Not in the usual sense. The Lebesgue measure of any countable (including any finite) set is zero.--Jasper Deng (talk) 05:51, 19 July 2017 (UTC)

But counting measure will do the trick.

By the way, the formula for computation you suggest (the sum of integrals) does not make sense. YohanN7 (talk) 07:29, 19 July 2017 (UTC)

Not for infinite sets it wouldn't.--Jasper Deng (talk) 07:36, 19 July 2017 (UTC)

Why not? Integrals are generally allowed to assume the values $\pm \infty$. In the present case, since strictly positive functions are considered, only finite values or $+\infty$ can occur, and the integral is never ill-defined. YohanN7 (talk) 08:25, 19 July 2017 (UTC)

I (personally) usually don't tend to consider an integral to exist with infinite value, rather that it exists or doesn't.--Jasper Deng (talk) 08:39, 19 July 2017 (UTC)

I have no problem with that. But even if the set is infinite, the integral may be finite. In the present case one has

$\int _{S^{*}}fd\mu =\Sigma _{x\in \mathbb {R} }x\cdot \mu (f^{-1}(x)).$

(So $f$ is actually a step function or simple function according to some definitions (e.g. Herbert Federer's Geometric measure theory) of the term. The seemingly uncountable sum needs its proper definition of course, see Federer again.) The r h s is a countable sum, with strictly postitive terms. It may well be finite. YohanN7 (talk) 09:02, 19 July 2017 (UTC)

Yes, it can indeed be finite since for some functions, $S^{*}$ can be chosen to produce an (absolutely) convergent series. But it won't work for every function and every $S^{*}$ meeting the OP's definition.--Jasper Deng (talk) 09:11, 19 July 2017 (UTC)

There's no point in arguing over conventions and terminology, but I dare say that in measure and integration theory, infinite values are allowed more often than not, both for the range of functions and the value of integrals. This is probably because it simpliefies the exposition. The only things not allowed (i.e. left undefined) are expressions of the type $\infty -\infty$. Whenever $0\cdot \infty$ occurs, it is defined to be zero (at least, I can't think of an exception atm). (Division may also cause problems, but rarely pops up.) YohanN7 (talk) 09:17, 19 July 2017 (UTC)

Yes, this is all very standard. Excluding the value ∞ serves no purpose. Measures take values in the range [0, ∞]. Integrals can be infinite, or they can fail to exist, and these are importantly different cases; excluding the value ∞ would collapse them together, to no benefit. --Trovatore (talk) 07:50, 23 July 2017 (UTC)

Questions: The OP says that "it appears that $\int _{D}f(x)dx=\sum _{k=1}^{n}\int _{x_{k}}^{x_{k}}f(x)dx=0$" but I wonder if this is meant to be $\int _{D}f(x)dx=\sum _{k=1}^{n-1}\int _{x_{k}}^{x_{k+1}}f(x)dx$ because for any constant $c\in (a,b)$ where f(x) is continuous and differentiable, $\int _{c}^{c}f(x)dx=0$. Also, is it meant to be that $a<x_{1}<x_{2}<...<x_{n}<b$? EdChem (talk) 07:41, 23 July 2017 (UTC)