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I cant find a counter example for a function that is Riemann integrable, but not Darboux integrable.

The only source i found on this forum is trying to prove the direction "If Riemann integrable, then Dargboux integrable". But i really doubt its truth.

Riemann integrability implies Darboux integrability

But as you can see, there is nothing interesting.

Charming Imp
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  • Look here https://en.wikipedia.org/wiki/Darboux_integral where it is said existence of the two integrals is equivalent and if they exist they have the same value. I didn't look through it to see if proofs included. – coffeemath Sep 07 '22 at 10:01
  • Ok, so for example we can have Volterras function which is not Riemann integrable and hence not Darboux integrable. But that conclusion we only know from Riemann integral sandwiched between Darboux sums. Hows about taking the function f(x) = x+2 if x is rational and 1 if x is irrational, and trying to improve it to a Riemann integrable so that near zero the integral is equivalent to darboux sums at least and is not equal at most. – Charming Imp Sep 07 '22 at 10:14
  • Your example of the last comment is not integrable in either sense, on any interval no matter how short. It cannot be modified to become integrable either unless modified so much it's not at all like the unmodified function. I mean you can take any function and "modify" it to make it e.g. constant, but that is not of interest. – coffeemath Sep 07 '22 at 10:18

1 Answers1

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They are indeed equivalent concepts.

Let's begging with the precise definitions:

A bounded function $f:[a,b] \rightarrow \mathbb{R}$ is Riemann integrable if there exists a real number $I$ such that given $\varepsilon>0$ there exists a $\delta>0$ such that for any partition $P=\{a=x_0 < x_1 < \cdots < x_n\}$ with diameter less than $\delta$, and any family of intermedian points $T$, with $t_i \in [x_{i-1}, x_{i}]$ with $i=1, \cdots, n$, we have that $$\left| \sigma(P,T) -I\right|<\varepsilon$$ where $$\sigma(P,T)=\sum_{i=1}^n f(t_i)(x_i-x_{i-1})$$ We say that $I$ is the integral of $f$ on $[a,b]$.

A bounded function $f:[a,b] \rightarrow \mathbb{R}$ is Darboux integrable if its superior and inferior integrals are the same, where these integrals are defined as $$\underline{\int_a^b}f = \sup\left\{ s(P): P \text{ is a partition of } [a,b] \right\}$$ $$\overline{\int_a^b}f = \inf\left\{ S(P): P \text{ is a partition of } [a,b] \right\}$$ Here, given a partition $P=\{a=x_0 < x_1 < \cdots < x_n\}$ it is defined $$s(P)=\sum_{i=1}^n \left(\inf_{[x_{i-1},x_i]}f(x)\right) (x_i-x_{i-1})$$ $$S(P)=\sum_{i=1}^n \left(\sup_{[x_{i-1},x_i]}f(x)\right) (x_i-x_{i-1})$$ We say that $I=\underline{\int_a^b}f=\overline{\int_a^b}f$ is the integral of $f$ on $[a,b]$.

Now, to prove the equivalence, we need some lemmas:

LEMMA 1: $f:[a,b] \rightarrow \mathbb{R}$ is Darboux integrable if, and only if, given $\varepsilon>0$ there exists a $\delta>0$ such that for any partition $P$ with diameter less than $\delta$ we have that $S(P)-s(P)<\varepsilon$.

LEMMA 2: Given a function $f:[a,b] \rightarrow \mathbb{R}$ and a partition $P$ of $[a,b]$ it is satisfied: $$s(P)=\inf\left\{\sigma(P,T): T \text{ is an intermedian family of points of }P \right\}$$ $$S(P)=\sup\left\{\sigma(P,T): T \text{ is an intermedian family of points of }P \right\}$$

With these two lemmas it is easy to prove the result:

First suppose that $f:[a,b] \rightarrow \mathbb{R}$ is Riemann integrable with integral $I$.

Then, given $\varepsilon>0$ there exists a $\delta>0$ such that for any given partition $P$ with diameter less than $\delta$, and any intermedian family of point $T$ we have that $|\sigma(P,T) - I|<\varepsilon/3$.

So, by Lemma 2, we have that if the diameter of $P$ is less than $\delta$ then $$I-\varepsilon/3 \leq s(P) \leq S(P) \leq I+\varepsilon/3 \hspace{5mm} (1)$$ ans so $$S(P)-s(P) \leq (I+\varepsilon/3) - (I-\varepsilon/3) < \varepsilon$$ Then, by Lemma 1 we have that $f$ is Darboux integrable (and furthermore, it is easy to see from (1) that $I$ is equal to both superior and inferior integrals).

On the other hand, suppose now that $f:[a,b] \rightarrow \mathbb{R}$ is Darboux integrable with integral $I$.

Then, given any partition $P$ of $[a,b]$ and any family of intermedian points $T$, we have by Lemma 2 that $$s(P) \leq \sigma(P,T) \leq S(P)$$ and by definition of Darboux integral that $$s(P) \leq I \leq S(P)$$ so we have that $$\left| \sigma(P,T) -I \right| \leq S(P)-s(P) \hspace{5mm} (2)$$ Now, given $\varepsilon>0$, by Lemma 1, there exists a $\delta>0$ such that for any given partition $P$ with diameter less than $\delta$ we have that $S(P)-s(P)<\varepsilon$.

Then, by (2) we have that if the diameter of $P$ is less than $\delta$ and $T$ is any family of intermedian points of $P$, then $$\left| \sigma(P,T) -I \right| \leq S(P)-s(P) < \varepsilon$$ and that implies, by definition, that $f$ is Riemann integrable.

EDIT

The proof of Lemma 2 is immediate from the definitions, and to prove Lemma 1 is enough to prove the following result:

Given a bounded function $f:[a,b] \rightarrow \mathbb{R}$, a partition $P_0$ of $[a,b]$ and $P$ another partition obtained by adjoining $p$ points to $P_0$, we have that $$s(P)-s(P_0) \leq 2pK|P|$$ $$S(P_0)-S(P) \leq 2pK|P|$$ where $K$ is any bound of $|f(x)|$ and $|P|$ denotes the diameter of $P$.

And, this one can be proven by induction on $p$.

Eparoh
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