# Countable and uncountable sets in Riemann integration

The Riemann integral over $[a,b]$ of a continuous function $f$ is $$\int\limits_a^bf(x)dx=\lim\limits_{\delta\rightarrow 0} \sum\limits_{i=0}^{n-1} (x_{i+1}-x_i)f(c_i)$$ where $c_i\in[x_i,x_{i+1}]$ and $\delta=\sup(x_{i+1}-x_i)$. The set $[a,b]$ is uncountable. When $\delta\rightarrow 0$, the number of elements in a set $[x_i,x_{i+1}]$ goes to $1$ and so the number of terms in the sum above is countable. How could a counatble set form a partition of an uncountable set (the sets $]x_i,x_{i+1}[$ form a partition of $]a,b[$).

Now I suspect that there's something wrong in "when $\delta\rightarrow 0$, the number of elements in a set $[x_i,x_{i+1}]$ goes to $1$" but even if I replaced it with "when $\delta\rightarrow 0$, the number of elements in a set $[x_i,x_{i+1}]$ goes to an integer $k$" or "when $\delta\rightarrow 0$, the number of elements in a set $[x_i,x_{i+1}]$ is countable" (I don't know why it isn't $1$) I still have the same problem.

• $\delta$ never actually reaches zero. We simply let it be as small (but positive) as we like to make the sum arbitrarily close to a limiting value.
– user169852
Commented Aug 18, 2014 at 19:38
• See the third comment under my answer. Commented Aug 18, 2014 at 20:57
• @JonasMeyer Let us take a uniform partition $x_{i+1}-x_i=\frac{b-a}{n}$. $\delta\rightarrow 0$ means $n\rightarrow\infty$ so the length of the intervals $I_{(k,n)}=[a+k\frac{b-a}{n},a+(k+1)\frac{b-a}{n}]$ goes to zero. What is the number of elements in $I_k=\lim\limits_{n\rightarrow\infty}I_{(k,n)}$?
– Kaku
Commented Aug 19, 2014 at 12:49
• Hafej, That question doesn't make sense to me. That limit isn't defined. Also, even if it were, what relevance would it have to finding a limit of Riemann sums, which is a number, not a set? Commented Aug 19, 2014 at 12:52
• @DaveL.Renfro You can't apply cantor's intersection theorem here since that's not a nested sequence: If $I_{(k,n)}=[x_k, x_{k+1}]$ where $k=0,1,\cdots,(n-1)$ then $I_{(k,n+1)}\subset I_{(k,n)}$ isn't true in general.
– Kaku
Commented Aug 25, 2014 at 16:07

The most simple limit sum to think of is $$\sum_{k=1}^n \frac 1n =1$$ as $1/n\to 0^+$. Even being that simple, it will clear up the situation. Staring at it, you might say

Well, in the limit $1/n$ goes to $0$ and there is a countable number of terms in this sum. There is clearly something wrong. How could a countable number of zeros sum up to one?

This question is more fundamental than integrals or Riemann sums, it is a question on the behaviour of limits. The formal definition of a limit will rescue us. To refresh our minds, a quick reminder: we say $\lim\limits_{x\to 0}f(x)=L$ if, and only if, for every $\varepsilon>0$ there is $\Delta >0$ such that $$0<|x|<\Delta\implies |f(x)-L|<\varepsilon.$$

Note that we never mentioned what would happen if $x=0$. It does not concern our definition, and that's the fundamental mistake here. We should not judge the sum in the hypothetical situation $1/n=0$, but only when $0<1/n - 0 <\Delta$ for some finite $\Delta$. The paradox disappears, because at any such interval, $\sum 1/n = 1$.

At any finite $\delta=\sup(x_{i+1}-x_i)$, the cardinality of $[x_i, x_{i+1}]$ is definitely not approaching $1$, even though it is $1$ when $\delta =0$. This illustrates a crucial difference between $\delta\to 0$ and $\delta =0$ that cannot be overestimated, and you will highly benefit from thinking about it. To finish the discussion, you might be interested in rereading some content on continuous functions.

• Well, Ian I agree with you. Your explanation is clear. I know the $\epsilon-\delta$ definitions of limits and continuity and it's clear that when $\delta$ is very small (not zero) $[x_i,x_{i+1}]$ isn't countable and there's no problem. My problem is what happens when $\delta=0$? Won't we have a countable union of countable (or finite) sets?
– Kaku
Commented Aug 26, 2014 at 9:24
• Hafej, just to make it clear: what happens when $\delta=0$ does not matter in any way for the limit. As we will see, the case $\delta=0$ is ill-defined and basically can do what we want it to do. If we insist that it is still a partition of the original interval, it cannot generate only countably many intervals, and for any $p\in[a,b]$ there is a set $\{p\}$ now. If we insist there are only countably many intervals, they cannot cover $[a,b]$ (and be a partition of it) because it would generate a countable set. Commented Aug 26, 2014 at 15:23
• We either have to give up the fact that the union of the intervals is a partition or that there are only countably many intervals. In my opinion, the second option is much more sensible. Commented Aug 26, 2014 at 15:24

I am not a mathematician, but from a scientist's perspective, going from the sum to the integral is going from discrete mathematics, where the variable is only defined over discrete values (such as where the physical quantity is a finite dimensional vector), to continuum mathematics, where the variable is defined over a continuum of values (the physical quantity is a function of a real or complex variable). So, I do sympathize with the question's author that in the discrete case, the sum contains the function evaluated at each point where the variable is defined but in the continuum case, the limit of the sum does not contain the function (integrand) evaluated at each of the continuum of points.

I think the key is to understand that not all sets that are countably infinity in size have the same properties. For instance, for the integers, there is such a thing as "next integer". However, for the rationals, which are also countably infinite, there is no "next rational". I think that the countably infinite ordered set of x values, for which the integrand is evaluated in the limit, are such that there is no "next x" in that ordered set. This allows $\delta$ to approach zero in the limit, without having to evaluate the integrand at each of the non-countably infinite set of values of x that are in the interval of integration.