# Inequality for Riemann-Stieltjes integral

Suppose $f$ is Riemann-Stieltjes integrable with respect to $g$. If $g$ is increasing then it is easy to show

$$\newcommand{\d}[1]{\,\mathrm{d}#1} \bigg| \int_a^bf(x)\d{g(x)} \bigg| \leq \int_a^b|f(x)| \d{g(x)}$$

just using $-|f| \leq f \leq |f|$ and the monotonicity of the integral.

Question: If $g$ has bounded variation but not monotone then the inequality is $$\bigg|\int_a^b f(x)\d{g(x)}\bigg| \leq \int_a^b|f(x)| \d{v_a^x(g)}$$

Here $v_a^x(g)$ is the total variation function. How can this be proved?

• I guess you can decompose $g=g_1-g_2$, where $g_1$ and $g_2$ are monotone increasing. Then $|\int f\, dg|=|\int fd g_1 - \int fdg_2|\le \int |f| d(|g_1|+|g_2|)$, and the latter term should equal $\int |f|d v_a^x(g).$ – Giuseppe Negro Jan 2 '18 at 19:03

\begin{align} \left|\int_{a}^{b}fdg\right| &= \lim_{\|\mathcal{P}\|\rightarrow 0}\left|\sum_{n=1}^{N}f(x_j^*)(g(x_j)-g(x_{j-1}))\right| \\ & \le \lim_{\|\mathcal{P}\|\rightarrow 0}\sum_{n=1}^{N}|f(x_j^*)||g(x_j)-g(x_{j-1})| \\ & \le \lim_{\|\mathcal{P}\|\rightarrow 0}\sum_{n=1}^{N}|f(x_j^*)|v_{x_{j-1}}^{x_j}(g) \\ & = \lim_{\|\mathcal{P}\|\rightarrow 0}\sum_{n=1}^{N}|f(x_j^*)|\{v_a^{x_j}(g)-v_a^{x_{j-1}}(g)\} \\ & = \int_{a}^{b}|f(x)|d_{x}v_{a}^{x}(g). \end{align}

• This is helpful (+1) and makes sense if the upper bound limit exists. But I am unclear how we know or show that the limit on the right-side exists and equals $\int_a^b |f| dv_a^x(g)$. – AlRacoon Jan 2 '18 at 22:52
• @AlRacoon The limit of the last sum on the right is the definition of the Riemann-Stieltjes integral of $|f|$ with respect to the monotone function $w(x)=v_a^x(g)$. Both integrals will exist if $f$ is integrable with respect to the variation of $g$, and then, also, the inequalities above will hold. – DisintegratingByParts Jan 2 '18 at 23:07
• My lingering question is if $f$ is integrable with respect to $g$ then is it always true that $f$ is integrable with respect to the variation? I know for example that a function could be integrable with respect one increasing integrator but not another depending on where discontinuities are located. I think I will accept this and ask another question. – AlRacoon Jan 3 '18 at 1:26

Indeed for an increasing function $f$ on $[a,b]$ one always has: $$\color{blue}{V^x_a (f) = f(x)-f(a)}$$

Then Note the following Property:

Another characterization states that the functions of bounded variation on a compact interval are exactly those $g:[a,b]\to \Bbb R$ which can be decomposed as $g= u_g -v_g$ where both $u_g$and $v_g$ are bounded and increasing functions.( that is $g$ is the difference of two increasing functions.) read the introduction here or here which more difficult

Moreover, indeed from definition we know that $$\color{red}{V^x_a (g)= u_g(x)-u_g(a) +v_g(x)-v_g(a)}\implies \color{blue}{dV^x_a (g)= du_g(x)+dv_g(x)}$$

therefore: $$\bigg|\int_a^b f(x)\d{g(x)}\bigg|= \bigg|\int_a^b f(x)\d{u_g(x)}- \int_a^b f(x)\d{v_g(x)}\bigg| \\\leq \bigg|\int_a^b f(x)\d{u_g(x)}\bigg|+\bigg| \int_a^b f(x)\d{v_g(x)}\bigg| \\\leq\int_a^b|f(x)| \d{u_g(x)}+\int_a^b|f(x)| \d{v_g(x)} \\\color{blue}{=\int_a^b|f(x)| d \left({u_g(x)+v_g(x)}\right)=\int_a^b|f(x)| \d{v_a^x(g)}}$$

Then you can conclude from the first case since you already done the proof for increasing function

• Thanks. I don't understand the relevance of the first statement because I said that $g$ is not monotone. I get the formal manipulations, but how do we know $f$ is integrable with respect to the variation. I have given no condition on $f$ like continuous other than integrable with respect to $g$? Finally how do we know that $dV = du_g + dv_g$? Those functions are not unique. – AlRacoon Jan 2 '18 at 21:10