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.
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.

  • 1
    $\begingroup$ not the $\max,$ the $\sup$ $\endgroup$ – zhw. Apr 5 '18 at 21:07

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.

  • $\begingroup$ Hi Lockjaw, thanks for your reply. So I was basically correct on my analysis. But based on what you said here, the purpose of the Darboux sums is so that it is does not depend on the properties of the interval. But in many numerical examples i have come across I see that they always choose a partition so that the max and min occur on the endpoints of the sub-interval. $\endgroup$ – Palu Apr 5 '18 at 18:52
  • $\begingroup$ I guess that in terms of computations they choose to construct and interval this way as to make the computation easier. $\endgroup$ – Palu Apr 5 '18 at 18:56
  • $\begingroup$ Also, i forgot to add that the function is always bounded on a finite closed interval. $\endgroup$ – Palu Apr 5 '18 at 18:57
  • $\begingroup$ HI @Palu, I'm happy to help! Indeed, the analysis goes better when you choose an interval in this way, however it's not a requirement. And yes, one purpose of the sum being constructed this way is that it doesn't depend on the way the partition "looks". I should also note that when you choose the intervals nicely you can often prove things like upper and lower bounds on the Darboux sums gracefully. This of course has implications for values that the integral can take on. $\endgroup$ – Lockjaw Apr 5 '18 at 18:58
  • $\begingroup$ Hi Lockjaw, thanks for these details. This is what I wanted to know. But now I feel this goes into another question, of how Darboux sum can become the Riemann sum. $\endgroup$ – Palu Apr 5 '18 at 19:03

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.

  • $\begingroup$ Hi Christian, thanks for your reply. Yes, I was under the impression that one would have to do N maximum problems for N-sub-intervals. But when you start talking about taking the differences, are you implying that taking my numerical example, one can do the computation more quickly just with the differences somehow? $\endgroup$ – Palu Apr 5 '18 at 19:02
  • $\begingroup$ Hi Christian, this is very interesting, I have not seen this approach that you are showing. But when you have U(f,P)-L(f,P) < epsilon, way this seems to be the start of trying to show that this leads to Riemann sums. $\endgroup$ – Palu Apr 5 '18 at 23:39

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