Constructing the Lebesgue-Stieltjes Integral as a Functional on $C^1 ([a,b])$ This is based on an approach to a homework problem from last year that I discarded but would like to fill in the details of now.  Unfortunately I can't see the necessary inequality.
Suppose we are given a function $g$ of bounded variation on $[a,b]$.  I would like to construct the Lebesgue-Stieltjes integral with respect to $g$ by defining a bounded linear functional on $C^1([a,b])$ by $\varphi \mapsto -\int_{\mathbb{R}} g(x) \varphi^{'}(x) dx$ then extending it to $C([a,b])$ by density and then applying the Riesz Representation Theorem.  To do this, however, I need to show that $|\int g(x) \varphi^{'}(x) dx|\leq C||\varphi||_\infty$ (notice that there is no derivative here--this is the condition to make it a bounded functional on $C^0([a,b])$) which is where the trouble comes in since I am not sure how to get such a bound without integrating by parts (which we can't do since $g$ need not be absolutely continuous).
I'm sure there is a nice way to handle this but I just don't see it.
Thanks for the help.
 A: Since you are working on $C^1([a,b])$, you can make use of the usual Riemann-Stieltjes integral to see that
$$
\begin{align*}
\int_a^b \phi(x) \, dg(x) &= \left[\phi(b)g(b) - \phi(a)g(a)\right] - \int_a^b g(x) \, d\phi(x) \\
&=  \left[\phi(b)g(b) - \phi(a)g(a)\right] - \int_a^b g(x)\phi'(x) \, dx
\end{align*}
$$
And using $\left|\int_a^b \phi(x) \, dg(x) \right|\le \|\phi\|_{\infty} \; V_a^b(g)$ (where $V_a^b(g)$ denotes the total variation of $g$), we obtain
$$\left|\int_a^b g(x)\phi'(x) \, dx\right| \le \left[|g(a)|+|g(b)|+V_a^b(g)\right] \, \| \phi\|_\infty = C\cdot \|\phi\|_\infty$$
So your functional is indeed continuous (and the Lebesgue-Stieltjes integral is an extension of the Riemann-Stieltjes integral).

In case you are not familiar with the Riemann-Stieltjes integral. The first equality can be seen as follows: Given a partition $P = \{x_0,x_1, \dots, x_N\}$ of $[a,b]$; $a= x_0 < x_1 <\dots < x_N=b$ and $\xi_i \in [x_{i-1}, x_{i}]$, we can write
$$
\begin{align}
\sum_{n=1}^N f(\xi_i) [g(x_i) - g(x_{i-1})] &= f(\xi_{N+1})g(b) - f(\xi_1)g(a) -  \sum_{n=1}^N \left[f(\xi_{i+1}) - f(\xi_i)\right] g(x_{i}) \\
&=  f(\xi_{N+1})g(b) - f(\xi_1)g(a) -  \sum_{n=1}^N \frac{f(\xi_{i+1}) - f(\xi_i)}{\xi_{i+1} - \xi_i} g(x_{i}) \left[\xi_{i+1} - \xi_i\right]
\end{align}
$$
setting $\xi_{N+1} = b$. Note that $$\left|\sum_{n=1}^N f(\xi_i) \left[g(x_i) - g(x_{i-1})\right]\right| \le \sum_{n=1}^N \left|f(\xi_i)\right| \, \left|g(x_i) - g(x_{i-1})\right|\le \|f\|_\infty \, V_a^b(g)$$ for any partition $P$.
So if $f \in C^1([a,b])$ and $g$ is of bounded variation, then both are Riemann-integrable and by taking the limit $|P|\to 0$ (finesse of the partition) we will obtain the estimate
$$\left| f(b)g(b) - f(a)g(a) - \int_a^b f'(x) g(x) \, dx \right| \le \|f\|_\infty \, V_a^b(g)$$
