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I am reading the Wikipedia explanation on Riemann Integration and how it explains the mesh needs to go to zero.

https://en.wikipedia.org/wiki/Riemann_integral

I have the following questions:

  1. When we say, the mesh goes to zero, does this ALSO imply the partitions are getting finer, meaning $n\rightarrow\infty$ or are they two distinct conditions?

  2. Why is it not enough to just to say the partitions become finer and finer (i.e. $n\rightarrow\infty$) and leave out the condition that the mesh eventually goes to zero? Or is it enough?

  3. If $n\rightarrow\infty$ and $mesh\rightarrow0$ are distinct concepts, how would you explain this? What is an example the former is met but the latter is not satisfied such that the Riemann integration is not a good approximation of the area under a curve?

  4. Is it the case $n\rightarrow\infty\implies mesh\rightarrow0?$ If not, why?

Frank Swanton
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    I first thought of if you have a sequence of partitions that is concentrating in one area, say, ${1/n}_{n=1}^{N}$ as a partition for $[0,1]$. Maybe this answers 2,3,4? If you say the mesh is going to $0$, then you should eventually be adding more points to your partition (even after just, say, moving your points around to be more equally distributed), and this would make $n\to \infty$. But the above example is one where $n\to \infty$ but the mesh is not going to $0$. – Keshav Feb 18 '20 at 23:04
  • See the difference between partitions getting finer and mesh tending to $0$ in this answer: https://math.stackexchange.com/a/2047959/72031 – Paramanand Singh Feb 19 '20 at 09:32
  • Btw partitions getting finer is not the same as $n\to\infty $ but rather adding points to the existing partition to make it finer. Not just the number of sub-intervals, but also their location matters in a partition. – Paramanand Singh Feb 19 '20 at 09:36

2 Answers2

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It occurs to me that you might be thinking about partitions $a=x_0 < x_1 < ... < x_n=b$ in which the partition intervals $[x_{i-1},x_i]$ have constant length $x_i - x_{i-1} = \frac{b-a}{n}$.

But it is important to note carefully that in the definition of a Riemann sum as given in that link, for a partition $a=x_0 < x_1 < ... < x_n = b$, the individual partition intervals $[x_{i-1},x_i]$ are not required to have constant length $x_i - x_{i-1} = \frac{b-a}{n}$. Because of this, it is quite possible to construct a sequence of partitions with $n \to \infty$ but mesh not going to zero. Thus, $n \to \infty$ is not a useful concept for this general class of partitions that are used for the Riemann integral.

In particular, even for an integrable function, suppose you have a sequence of tagged partitions with $n \to \infty$ but whose mesh does not go to zero; think of the partition $0 < 1/2 < 1$, then subdivide the interval $[0,1/2]$ more and more finely, but don't subdivide the interval $[1/2,1]$ at all. In this situation there is little expectation that the Riemann sums will converge, nor, if they do, that they will converge to the value of the Riemann integral.

Lee Mosher
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To summarize some of the other responses: The condition "mesh goes to zero" is stronger than the condition "$n \rightarrow \infty$", i.e. the zero-mesh condition implies the number of intervals condition. [Proof: if we never had more than $N$ intervals, then one of our intervals would always be at least $(b-a)/N$ in length, so the mesh couldn't go to zero.]

And the zero-mesh condition is strictly stronger than the $n \rightarrow \infty$ condition, as you can come up with situations where the number of intervals goes to infinity but the mesh doesn't go to zero (e.g. LeeMosher's answer).

BTW, when I was first learning Riemann integrals I thought that writing them with $\lim_{\| S \| \rightarrow 0}\sum$ notation (where $\|S\|$ was the mesh of a partition) was pretty bogus. It's not like the Riemann sum is a direct function of the mesh of the partition, so it's not really like when we did $\lim_{x \rightarrow a} f(x)$. I just told myself that we were using the same notation for two different but similar concepts. It wasn't until I took topology and learned about limits over nets, and how they could be used to define integrals that I could see them as two instances of a single definition.

JonathanZ
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