### Riemann-, Generalized-Riemann-, and Darboux-Stieltjes integrals

(The following is somewhat rough and may have typos.)

Let us begin by setting the notations and recalling what happens without the Stieltjes part.

Defn (Partition)
Let $I$ be a closed interval. A partition $P$ is a finite collection of closed subintervals $\{I_\alpha\}$ such that

1. $P$ is finite;
2. $P$ covers $I$, i.e. $\cup P = I$;
3. $P$ is pairwise almost disjoint, i.e. for $I_\alpha, I_\beta$ distinct elements of $P$, their intersection contains at most one point.

We write $\mathscr{P}$ for the set of all partitions of $I$.

Defn (Refinement)
Fix $I$ a closed interval, and $P, Q$ two partitions. We say that $P$ refines $Q$ or that $P \preceq Q$ if for every $I_\alpha\in P$ there exists $J_\beta \in Q$ such that $I_\alpha \subseteq J_\beta$.

Defn (Selection)
Given $I$ a closed interval and $P$ a partition, a selection $\sigma: P \to I$ is a mapping that satisfies $\sigma(I_\alpha) \in I_\alpha$.

Defn (Size)
Given $I$ a closed interval and $P$ a partition, the size of $P$ is defined as $|P| = \sup_{I_\alpha \in P} |I_\alpha|$, where $|I_\alpha|$ is the length of the closed interval $I_\alpha$.

Remark In the above we have defined two different preorders on the set $\mathscr{P}$ of all partitions. One is induced by the size: we say that $P \leq Q$ if $|P| \leq |Q|$. The other is given by the refinement $P\preceq Q$. Note that neither are partial orders. (But that the preorder given by refinement can be made into a partial order if we disallow zero-length degenerate closed intervals.) Note also that if $P\preceq Q$ we must have $P \leq Q$.

Now we can define the notions of integrability.

Defn (Integrability)
Let $I$ be a closed, bounded interval and $f:I \to \mathbb{R}$ be a bounded function. We say that $f$ is integrable with integral $s$ in the sense of

• Riemann if for every $\epsilon > 0$ there exists $P_0\in \mathcal{P}$ such that for every $P \leq P_0$ and every selection $\sigma:P \to I$ we have
$\displaystyle \left| \sum_{I' \in P} f(\sigma(I')) |I'| - s \right| < \epsilon$

• Generalised-Riemann if for every $\epsilon > 0$ there exists $P_0 \in \mathcal{P}$ such that for every $P \preceq P_0$ and every selection $\sigma: P\to I$ we have
$\displaystyle \left| \sum_{I' \in P} f(\sigma(I')) |I'| - s \right| < \epsilon$

• Darboux if
$\displaystyle \inf_{P\in\mathscr{P}} \sum_{I' \in P} (\sup_{I'} f )|I'| = \sup_{P\in\mathscr{P}} \sum_{I' \in P} (\inf_{I'} f )|I'| = s$

From the definition it is clear that “Riemann integrable” implies “Generalised-Riemann integrable”. Furthermore, we have clearly that for a fixed $P$
$\displaystyle \sum_{I' \in P} (\inf_{I'} f) |I'| \leq \sum_{I' \in P} f(\sigma(I')) |I'| \leq \sum_{I' \in P} (\sup_{I'} f) |I'|$
and that if $P \preceq Q$ we have
$\displaystyle \sum_{I' \in Q} (\inf_{I'} f) |I'| \leq \sum_{I' \in P} (\inf_{I'} f) |I'| \leq \sum_{I' \in P} (\sup_{I'} f) |I'| \leq \sum_{I' \in Q} (\inf_{I'} f) |I'|$
so “Darboux integrable” also implies “Generalised-Riemann integrable”. A little bit more work shows that “Generalised-Riemann integrable” also implies “Darboux integrable” (if the suprema and infima are obtained on the intervals $I'$, this would follow immediately; using the boundedness of the intervals we can find $\sigma$ such that the Riemann sum approximates the upper or lower Darboux sums arbitrarily well.

The interesting part is the following
Theorem
Darboux integrable functions are Riemann integrable. Thus all three notions are equivalent.

Proof. Let $P, Q$ be partitions. Let $|P| \leq \inf_{I'\in Q, |I'| \neq 0} |I'|$, and let $m$ be the number of non-degenerate subintervals in $Q$. We have the following estimate
$\displaystyle \sum_{I'\in Q} (\inf_{I'} f) |I'| - (m-1) |P| (\sup_I 2|f|) \leq \sum_{J'\in P} f(\sigma(J')) |J'| \leq \sum_{I'\in Q} (\sup_{I'} f) |I'| + (m-1) |P| (\sup_I 2|f|)$
The estimate follows by noting that “most” of the $J'\in P$ will be proper subsets of $I'\in Q$, and there can be at most $m-1$ of the $J'$ that straddles between two different non-degenerate sub-intervals of $Q$. To prove the theorem it suffices to choose first a $Q$ such that the upper and lower Darboux sums well-approximates the integral. Then we can conclude for all $P$ with $|P|$ sufficiently small the Riemann sum is almost controlled by the $Q$-Darboux sums. Q.E.D.

Now that we have recalled the case of the usual integrability. Let us consider the case of the Stieltjes integrals: instead of integrating against $\mathrm{d}x$, we integrate against $\mathrm{d}\rho$, where $\rho$ is roughly speaking a “cumulative distribution function”: we assume that $\rho:I \to \mathbb{R}$ is a bounded monotonically increasing function.

The definition of the integrals are largely the same, except that at every step we replace the width of the interval $|I'|$ by the diameter of $\rho(I')$, i.e. $\sup_{I'} \rho - \inf_{I'} \rho$. The arguments above immediately also imply that

• “Riemann-Stieltjes integrable” implies “Generalised-Riemann-Stieltjes integrable”
• “Darboux-Stieltjes integrable” implies “Generalised-Riemann-Stieltjes integrable”
• “Generalised-Riemann-Stieltjes integrable” implies “Darboux-Stientjes integrable”

However, Darboux-Stieltjes integrable functions need not be Riemann-Stieltjes integrable. The possibility of failure can be seen in the proof of the theorem above, where we used the fact that $|P|$ is allow to be made arbitrarily small. The same estimate, in the case of the Stieltjes version of the integrals, has $|P|$ replaced by $\sup_{J'\in P} (\sup_{J'} \rho - \inf_{J'} \rho)$, which for arbitrary partitions need to shrink to zero. To have a concrete illustration, we give the following:

Example
Let $I = [0,1]$. Let $\rho(x) = 0$ if $x < \frac12$ and $1$ otherwise. Let $f(x) = 0$ if $x \leq \frac12$ and $1$ otherwise. Let $Q_0$ be the partition $\{ [0,\frac12], [\frac12,1]\}$. We have that
$\displaystyle \sum_{I'\in Q_0} (\sup_{I'} f) (\sup_{I'} \rho - \inf_{I'} \rho) = 0 \cdot (1 - 0) + 1\cdot (1 - 1) = 0$
while
$\displaystyle \sum_{I'\in Q_0} (\inf_{I'} f) (\sup_{I'} \rho - \inf_{I'} \rho) = 0 \cdot (1-0) + 0 \cdot(1-1) = 0$
so we have that in particular the pair $(f,\rho)$ is Darboux-Stieltjes integrable with integral 0. However, let $k$ be any odd integer, consider the partition $P_k$ of $[0,1]$ into $k$ equal portions. Depending on the choice of the selection $\sigma$, we see that the sum can take the values
$\displaystyle \sum_{I'\in P_k} f(\sigma(I')) (\sup_{I'} \rho - \inf_{I'}\rho) = f(\sigma([\frac12 - \frac1{2k},\frac12 + \frac1{2k}])) (1 - 0) \in \{0,1\}$
which shows that the Riemann-Stieltjes condition can never be satisfied.

The example above where both $f$ and $\rho$ are discontinuous at the same point is essentially sharp. A easy modification of the previous theorem shows that
Prop
If at least one of $f,\rho$ is continuous, then Darboux-Stieltjes integrability is equivalent to Riemann-Stieltjes integrability.

Remark The nonexistence of Riemann-Stieltjes integral when $f$ and $g$ has shared discontinuity points is similar in spirit to the idea in distribution theory where whether the product of two distributions is well-defined (as a distribution) depends on their wave-front sets.