TITLE: Characterizing BV functions via the Stieltjes integral
QUESTION [8 upvotes]: Let $J=[a,b]$ be a compact interval and $f,g:J\rightarrow\mathbb{R}$ be two bounded functions. We say that $\int f dg$ exists (and equals $A$) if and only if the following condition holds: for any $\epsilon>0$, there exist a partition $P_\epsilon$ of $J$ such that for any refinement $P\supset P_\epsilon$, any Riemann-Stieltjes sum with repect to $(f,P,g)$ lies in the $\epsilon$-neighborhood of $A$. 
In Hildebrandt, T. H. (1938), "Definitions of Stieltjes Integrals of the Riemann Type", The American Mathematical Monthly, 45 (5): 265–278, it is stated without proof that 

If $\int f dg$ exists for every $f(x)$ continuous on $a\leq x\leq b$, then $g(x)$ must be of bounded variation.


How do we prove such a statement? Should we use piece-wise linear functions alternating between 0 and 1 resembling step functions?? I don't see how I should proceed.. any advice would be welcome.

REPLY [9 votes]: This is a straightforward application of the Uniform Boundedness Theorem.  
For any partition $P = (x_0,x_1, \ldots,x_n)$ of $[a,b]$ define the linear operator $T_P:C([a,b]) \to \mathbb{R}$ by
$$T_Pf = \sum_{j=1}^nf(x_j)(g(x_j) - g(x_{j-1}).$$
Here $C([a,b])$ is taken to be the Banach space equipped with the supremum norm $\|f\|_\infty.$ 
Since every $f$ is integrable with respect to $g$, we have 
$$\lim_{\|P\| \to 0} T_Pf = \int_a^b f \, dg.$$
and, hence, for every $f$ there exists $M_f > 0$ such that $\sup_P|T_P f| \leqslant M_f$.
By the Uniform Boundedness Theorem we have uniform boundedness in operator norm and 
$$\sup_{f \in C([a,b]), \|f\|_\infty = 1} \sup_P|T_P f| \leqslant M < \infty.$$
For any partition $P$ we can construct a continuous (piecewise linear function) $f$ such that $f(x_j) = 1 $ if $g(x_j) - g(x_{j-1}) \geqslant 0$ and $f(x_j) = -1$ if $g(x_j) - g(x_{j-1}) < 0.$ Thus
$$\sum_{j=1}^n |g(x_j) - g(x_{j-1})| \leqslant M < \infty$$
and $g$ must be of bounded variation.