A question about general Riemann sums

Question

My Calculus textbook spent a good chunk of the chapter deriving Left/Right Riemann sums, only to ditch them for general Riemann sums, which they never bother to derive or illustrate.

I'm fairly confident that I understand the notion of a right (or left) Riemann sum. Let's say we have a partition $P$, of closed interval $[a,b]$, with subintervals $[x_k, x_k+1]$ (of equal length, of course). If we want to approximate the area under a given curve (on interval $[a,b]$) , we can multiply $Δx_k = (b-a)/n$ (the kth subinterval) by $f( x̄ )$ (the minimum y value on the kth subinterval) then take the sum of $f( x̄ )*Δx_k$ from $k = 1$ to $k = n$. And to obtain a better approximation, we can take the limit of this expression as $n$ goes to infinity. This method of approximation is not only geometrically, but also algebraically, obvious to me.

The mathematical clarity breaks down for me when we start talking about "general" Riemann sums. When in regards to general Riemann sums, the author states: "Since $Δx_k$ can now vary, it is now no longer enough to require that $n$ approaches infinity; we must also require that the length of the longest subinterval approaches zero. Since the latter condition includes the former, we now let "max $x_k$" denote the length of the longest subinterval, and take the sum (from $k = 1$ to $k = n$) of $f(x̄ )$ $Δx_k$ as $\max x_k$ approaches zero."

My question is, where does this relationship come from? With right Riemann sums we had the equation $Δx_k = (b-a)/n$ for each subinterval. This makes it perfectly clear that as $Δx_k$ approaches zero, n must approach infinity (because b-a is constant). For the general Riemann sums this equation doesn't hold, due to the fact that the subintervals vary in relative size, right? So what equation explicitly shows the relationship that the author claims? Namely that "the latter includes the former" in the case of a general Riemann sum?

Answer 1

Notice that as subintervals shrink the designation of 'longest subinterval' won't always apply to the same subinterval. So, as you keep shrinking the subinterval which is currently the longest you will also need to keep adding more subintervals. So, by making sure that you always take the length of the longest subinterval to zero you are also ensuring that the number of subintervals approaches infinity.

Answer 2

Suppose I have a definite integral on a closed interval $[A,B]$ where $A < B$ and I'm trying to compute the Riemann integral of a real-valued function $f$.

$$ \text{We compute the value of } \int_A^B f(x) \text{d}x $$

You have already mentioned the notion of a partition $P$. Let's make it a bit more formal. A set of closed intervals $W$ is a partition if and only if:

• The union of all elements of $W$ is a closed interval, no running off to infinity or leaving gaps.
• The intersection of any distinct pair of elements in $W$ is either a single point or the empty set.

Suppose we have a function $L$ from the closed real intervals to the real numbers, we can then define the following expression as our Riemann sum. We also constrain $L$ to only return a real number inside the interval given to it as an argument.

$$ \sum_{x \in P} f(L(x)) \cdot \|x\| \;\;\;\text{where $\|x\|$ is the length of the interval $x$} $$

In the case of a left Riemann sum, $L$ is just the minimum function, in the case of a right Riemann sum, $L$ is the maximum function. You also mention a different kind of Riemann sum where $L$ is the function that minimizes or maximizes the value of $f$ on the given interval. This is also a legitimate type of Riemann sum, although it is not a left or right Riemann sum.

When talking about a general Riemann sum, $L$ is arbitrary. It is constrained to always return a real number inside the interval that was given, but there are no further restrictions placed on $L$.

Next let's imagine an infinite sequence of partitions of $[A, B]$ and call this sequence $S$.

When taking the limit of left and right Riemann sums on partitions $[0,1]$ with evenly spaced subpartitions, we imagine something like the following on $S$.

$$ \left\{\left[0, 1\right]\right\} \;\text{then}\; \left\{\left[0, \frac{1}{2}\right], \left[\frac{1}{2}, 1\right]\right\} \;\text{then}\; \left\{\left[0, \frac{1}{3}\right], \left[\frac{1}{3}, \frac{2}{3}\right], \left[\frac{2}{3}, 1\right]\right\} \;\text{then}\; \cdots $$

However, when talking about a general Riemann sum, we don't prescribe an $S$, but rather impose some conditions on it.

Your book mentions two conditions that are supplied simultaneously. I will list them below and throw in a third constraint to rule out length zero closed intervals like $[3.2, 3.2]$.

• The number of items $S_k$ is $k$
• The limit of the length of the longest item in $S_k$ approaches zero as $k$ approaches infinity.
• None of the subintervals have length zero.

I think it's easier to see why the second condition is thrown in there by picking a bad value for $S$ that will mis-evaluate the definite integral.

[0,   1]
[0, 0.9], [0.9, 1]
[0, 0.9], [0.9, 0.99], [0.99, 1]
[0, 0.9], [0.9, 0.99], [0.99, 0.999], [0.999, 1]
and so on

The length of the longest partition stabilized immediately at $0.9$. We can rule out pathological situations like this by insisting that the length of the maximum segment approaches zero.

So, this gives us the following definition for the Riemann integral.

$$ \int_A^B f(x) \text{d}x = \xi \stackrel{\text{def}}{\iff} \xi = \lim_{k \to \infty} \sum_{x \in S_k} f(L(x)) \cdot \|x\| \;\;\text{for all choices of $S$ and $L$} $$

I will now make two further remarks that might be helpful.

1) We can simplify the conditions imposed on $S$.

We can instead impose only the second condition and third condition on $S$ and only require that the maximum length of a subpartition approaches zero. By forcing the length of the longest subpartition to get smaller and smaller, we necessarily increase the number of subpartitions.

2) The rational indicator function is not Riemann integrable

The rational indicator function, which we will write as $1_{\mathbb{Q}}$ returns $1$ if the argument it is given is rational and $0$ is the arugment it is given is irrational.

So, $1_{\mathbb{Q}}(\pi)$ is zero, but $1_{\mathbb{Q}}(4.5)$ is one.

However, if you imagine taking a closed interval on the real line of any nonzero length, that interval always contains both rational and irrational numbers.

Therefore your choice of $L$ is always relevant. In every single interval, you will have to decide whether to pick a rational or irrational number, so there's no guarantee that your sequence of Riemann sums will ever converge to anything.

In fact, the situation is worse than that. If you always pick an irrational number in every interval, the limit of the Riemann sums is $0$, but if you always pick a rational number in every interval, the limit of the Riemann sums is the length of $[A, B]$.