Is it necessary that if a function is Riemann integrable, then it is also Darboux integrable?

Question

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.

Answer 1

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 $p$.