Darboux sums : computations of Lower and Upper sums

Question

I have been having difficulty finding a good understanding of the computation of Darboux sums. I have seen other here ask similar questions, but it does not get to the core of the idea in my opinion.

If I have a partition, that is uneven, which is apparently allowed in Darboux sums, like partition of [0,7] into : {0, 0.5,3,4,5,5.5, 7}. And additionally say that we have a function that is NOT monotonic over this interval of [0,7].

My question is:
For each sub-interval, to find the upper sum, we try to find the MAX of the function over each sub-interval, which means that the MAX for the sub-interval can be found then on either the right or left end-points of the sub-interval, or we may need to find the max somewhere in between(which would require the use of derivative to find the max critical value.
OR
Does one need to choose a partition in a particular way such that over the particular sub-interval the function is Monotonic, hence one would construct the rectangles based on only the end-points to obtain either Upper or Lower sums.

Would this be the correct way of thinking about Darboux sums.

Hope someone can clarify.

Answer 1

As per the definition of the $M_i$ (whose weighted sum constitute the upper sum you mention) in the Wikipedia page, it appears that both parts of your "OR" statement are missing the mark. I think the intuition you have in your first statement is correct, but there are some caveats:

The main issue is that the quantity we are looking for is the supremum of the function over the interval (which can occur at either an endpoint, or somewhere in the middle of the interval). Since $f$ is a assumed to be a bounded function in the Wikipedia article one can conclude that the supremum always exists so the $M_i$ quantity is well defined. Actually finding $M_i$ can be a challenge if the function isn't well behaved, and as you mention could require some calculus. In the case of something like the Dirichlet function, the $M_i$ will be determined entirely by the existence of rational numbers in the sub-intervals of the partition we've chosen.

The issue with your second statement is that the partition is supposed to be arbitrary. That is, we don't want to have the existence of our upper sum depend on the properties of the partition (such as a max being obtained at an endpoint) -- our sum should "make sense" mathematically regardless of how we've chosen the partition. By using the supremum over a closed and bounded interval we obtain a construction which always makes sense provided $f$ is a bounded function.

Answer 2

Given a partition $$P:\qquad a=x_0<x_1<x_2<\ldots<x_N=b\tag{1}$$ of the interval $[a,b]$ the definition $$U(f,P):=\sum_{k=1}^N \overline{ f_k}\>(x_k-x_{k-1}),\qquad \overline{f_k}:=\sup\{f(x)\,|\,x_{k-1}\leq x\leq x_k\},$$ seems to require you to solve $N\gg1$ maximum problems in order to compute the exact value of $U(f,P)$ (you cannot assume that $f$ is monotone on each subinterval). But in reality you don't need this exact value, only estimates of the form $$U(f,P)-L(f,P)\leq\ldots\quad .$$ Such estimates come from information about differences $|f(x)-f(x')|$ when $|x-x'|$ is small.

Instead of Darboux sums (also called upper and lower Riemann sums) I suggest the following completely finitary setup: An ${\mathbb R}^n$-valued function $f$ is integrable over $[a,b]$ if for all $\epsilon>0$ there are a parttion $(1)$ of $[a,b]$ and bounds $\Delta_k\geq0$ such that $$|f(x)-f(x')|\leq \Delta_k\qquad\forall\> x,\> x'\in[x_{k-1},x_k]$$ and $$\sum_{k=1}^N\Delta_k\>(x_k-x_{k-1})<\epsilon\ .\tag{2}$$ The sum in $(2)$ is the precise stand-in for $U(f,P)-L(f,P)$, but does not involve any sups and infs.