I'm studying the Darboux definition of integrability, which I completely explained here. There's an exercise that asks me to prove that the Darboux Integrability is equivalent to Riemann Integrability, but this Riemann integral is defined as the following:

It first defines a 'pointed' partition (I don't know how to say it in ensligh) by the following: a 'pointed' partition $[a,b]$ is a pair $P^*=(E,ξ)$, where $P=\{t_0, t_1, \cdots, t_n\}$ is a partition of $[a,b]$ and $ξ = (ξ_1, \cdots, ξ_n)$ is a list of $n$ chosen numbers such that $t_{i-1}\le ξ_i\le t_i$ for each $i=1,\cdots ,n$.

Now, the Riemann Integral is defined as:

$$\sum(f,P^*) = \sum_{i=1}^n f(ξ_i)(t_i-t_{i-1})$$

(I didn't understand the notation for the left hand side of the equation, by the way)

Finally, I'm asked to prove the following: given $f,g:[a,b]\to \mathbb{R}$ integrable functions, for the entire partition $P=\{t_0, \cdots, t_n\}$ of $[a,b]$ let $P^* = (P,ξ)$ and $P^{\#} = (P, η)$ be pointed partitions of $P$, then:

$$\lim_{|P|\to 0}\sum f(ξ_i)g(η_i)(t_i-t_{i-1}) = \int_a^b f(x)g(x) \ dx$$

I guess here I need to prove that the Riemann Integral of the product of two functions if the darboux integral of $f(x)g(x)$, but it seems too obvious, I just need to verify that $f,g$ are integrable, then their product is too, isn't it? I'm pretty sure this should be a hard question. Is there another interpretation that I'm missing?

  • $\begingroup$ The notation $\sum (f, P^{*})$ is called a Riemann sum for $f$ over a pointed/tagged partition $P$ of $[a, b]$ and Riemann integral is then defined as the limit of such sums as the norm $|P|$ of the partition $P$ tends to $0$. Most books also use the notation $S(f, P^{*})$ instead of $\sum(f, P^{*})$. Note that this is the original definition of integral given by Bernhard Riemann. $\endgroup$
    – Paramanand Singh
    Commented Sep 28, 2016 at 8:04

1 Answer 1


I have discussed the equivalence of definitions of Riemann integral based on Darboux sums and Riemann sums in this and this answer.

The idea is that the Riemann sums are technically difficult to handle because of the arbitrary nature of $\xi_{i}$ in sub-interval $[t_{i - 1}, t_{i}]$ of a partition $T = \{t_{0}, t_{1}, t_{2}, \ldots, t_{n}\}$ of $[a, b]$. Hence the Darboux sums are used which replace the $f(\xi_{i})$ with supremum and infimum of $f$ on $[t_{i - 1}, t_{i}]$ and thereby remove dependency on the arbitrary $\xi_{i}$.

Now I come to your question. Suppose $f, g$ are Riemann integrable on $[a, b]$. Then it is possible to prove that their product $fg$ is also Riemann integrable on $[a, b]$. By definition of Riemann integral as a limit of Riemann sums we see that $$\int_{a}^{b}f(x)g(x)\,dx = \lim_{|P| \to 0}\sum_{i = 1}^{n}f(\xi_{i})g(\xi_{i})(t_{i} - t_{i - 1})\tag{1}$$ where the notation is taken from your question. You are now asked to prove that the points $\xi_{i}$ need not necessarily chosen to be same for $f$ and $g$ so that we can use $\xi_{i}$ for $f$ and another set of points $\eta_{i}$ for $g$. This is correct and we prove it below.

Note that \begin{align} |\Delta| &= \left|\lim_{|P| \to 0}\sum_{i = 1}^{n}f(\xi_{i})g(\xi_{i})(t_{i} - t_{i - 1}) - \lim_{|P| \to 0}\sum_{i = 1}^{n}f(\xi_{i})g(\eta_{i})(t_{i} - t_{i - 1})\right|\notag\\ &= \left|\lim_{|P| \to 0}\sum_{i = 1}^{n}f(\xi_{i})\{g(\xi_{i}) - g(\eta_{i})\}(t_{i} - t_{i - 1})\right|\notag\\ &\leq M\lim_{|P| \to 0}\sum_{i = 1}^{n}|g(\xi_{i}) - g(\eta_{i})|(t_{i} - t_{i - 1})\notag\\ &\leq M\lim_{|P| \to 0}\sum_{i = 1}^{n}\{M_{i}(g) - m_{i}(g)\}(t_{i} - t_{i - 1})\notag\\ &= M\lim_{|P| \to 0}\{U(P, g) - L(P, g)\}\tag{2}\\ &= M \cdot 0\notag\\ &= 0\notag \end{align} where $M$ is some bound for $|f|$ on $[a, b]$ and $M_{i}(g)$ and $m_{i}(g)$ are supremum and infimum of $g$ on $[t_{i - 1}, t_{i}]$. Hence $\Delta = 0$ and it follows that $$\int_{a}^{b}f(x)g(x)\,dx = \lim_{|P| \to 0}\sum_{i = 1}^{n}f(\xi_{i})g(\xi_{i})(t_{i} - t_{i - 1}) = \lim_{|P| \to 0}\sum_{i = 1}^{n}f(\xi_{i})g(\eta_{i})(t_{i} - t_{i - 1})\tag{3}$$

The limit in $(2)$ is $0$ and it is somewhat difficult to prove that it is $0$. I have proved it in the linked answer as the following theorem:

Integrability Condition 2 A function $f$ bounded on $[a, b]$ is Riemann integrable over $[a, b]$ if and only if for every $\epsilon > 0$ there is a number $\delta > 0$ such that $$U(P, f) - L(P, f) < \epsilon$$ whenever norm $|P| < \delta$.

Note: The result in question is called Bliss' Theorem which was proved by G. A. Bliss (see his paper) as an alternative to a more general result called Duhamel Principle.

  • $\begingroup$ I'm confused about your usage of the limit notation, $\lim\limits_{ |P| \to 0^+}$ (I've taken the liberty to assume that you meant the right hand limit as $|P|>0$). In any case, how does that limit make sense? $P$, a sequence, is not a function of $|P|$, positive real number (since two different partitions can have the same norm) so shouldn't $\lim\limits_{|P|\to 0^+} U(P,f)= U(P,f)$? $\endgroup$
    – William
    Commented Feb 1, 2022 at 11:25
  • $\begingroup$ @William: please see the definition given in math.stackexchange.com/a/2047959/72031 $\endgroup$
    – Paramanand Singh
    Commented Feb 1, 2022 at 21:15

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