Prove the Cauchy Criterion for Riemann-Stieltjes Integrability

Question

Suppose $f$ and $\alpha$ are bounded, real-valued functions on $[a,b]$.

Prove: $f \in R(\alpha)$ on $[a,b]\iff \forall \epsilon > 0, \exists P_\epsilon \in \mathcal{P} [a,b]$ such that, for all partitions $P$ and $P'$ finer than $P_\epsilon$, and all Riemann-Stieltjes sums $S(P,f,\alpha)$ and $S(P',f,\alpha)$, \begin{equation*} |S(P,f,\alpha)-S(P',f,\alpha)|<\epsilon \end{equation*}

(Notation: $f \in R(\alpha)$ means f is Riemann-Stieltjes integrable with respect to $\alpha$. $\mathcal{P} [a,b]$ means the set of all partitions on $[a,b]$.)

Attempt:

($\Rightarrow$)

In the forward direction ($f \in R(\alpha)$) we have: $\forall \epsilon > 0, \exists P_\epsilon$ such that \begin{equation} |S(P,f,\alpha)-\int _{a}^{b} fd\alpha| + |\int _{a}^{b} fd\alpha-S(P',f,\alpha)|<\epsilon/2 + \epsilon/2 = \epsilon \end{equation} \begin{equation} \Rightarrow |S(P,f,\alpha)-S(P',f,\alpha)|<\epsilon \text{ by triangle inequality} \end{equation}

($\Leftarrow$)

For the converse: $|S(P,f,\alpha)-S(P',f,\alpha)|<\epsilon$ implies that $\forall m \in \mathbb{N}$ there is a partition $P_m$ such that for all $P, P'$ finer than $P_m$ we have that $|S(P,f,\alpha)-S(P',f,\alpha)|<\frac{1}{m}$.

I'm not sure how to progress from here.

Answer 1

The proof of the converse is slightly clumsy and is independent of the nature of integrator function $\alpha$ and hence we will drop the symbol $\alpha$ and just use $S(f, P)$ in place of $S(f, P, \alpha)$. Moreover in what follows it is necessary to deal with the tags for the partition $P$ and use the notation $T_{P}$ for the a set of tags related to partition $P$. If we have two different tags for partition $P$ we can use notation $T_{P}$ and $T_{P}'$. So the usual notation for Riemann-Stieltjes sums will be $S(f,P, T_{P})$.

Thus we are given that for any $\epsilon > 0$ there is a partition $P_{\epsilon}$ of $[a, b]$ such that $$|S(f, P, T_{P}) - S(f, P', T_{P'})| < \epsilon$$ for all $P, P'$ finer than $P_{\epsilon}$. Choosing $\epsilon = 1/n$ for positive integer $n$ we see that there is a partition $P_{n}$ such that $$|S(f, P, T_{P}) - S(f, P', T_{P'})| < \frac{1}{n}$$ for all $P, P'$ finer than $P_{n}$. Clearly we can replace $P_{n + 1}$ by $P_{n}\cup P_{n + 1}$ and hence it is OK to assume that $P_{n}\subseteq P_{n + 1}$.

Consider a sequence $a_{n} = S(f, P_{n}, T_{P_{n}})$ for some choice of tags for $P_{n}$. A different choice of tags will lead to a different sequence $b_{n} = S(f,P_{n}, T_{P_{n}}')$. Clearly if $m > n$ then we have $|a_{m} - a_{n}| < 1/n$ and hence $a_{n}$ is a Cauchy sequence (and so is $b_{n}$). Let $I = \lim_{n \to \infty}a_{n}$. We will prove that $\int_{a}^{b}f\,d\alpha = I$.

Clearly letting $m \to \infty$ in the inequality $|a_{m} - a_{n}| < 1/n$ we get $|a_{n} - I| \leq 1/n$. Now let $\epsilon > 0$ be given and we choose a positive integer $n > 2/\epsilon$. Let $P$ be a partition of $[a, b]$ with $P \supseteq P_{n}$ and $T_{P} $ be any choice of tags associated with partition $P$. Then we have \begin{align} |S(f, P, T_{P}) - I| &= |S(f, P, T_{P}) - S(f, P_{n}, T_{P_{n}}) + S(f, P_{n}, T_{P_{n}}) - I|\notag\\ &\leq |S(f, P, T_{P}) - S(f, P_{n}, T_{P_{n}})| + |S(f, P_{n}, T_{P_{n}}) - I|\notag\\ &< \frac{1}{n} + \frac{1}{n}\notag\\ &< \epsilon\notag \end{align} It now follows that $f\in R(\alpha)$ and $\int_{a}^{b}f\,d\alpha = I$.