# Riemann-Stieltjes Integral with respect to total variation

In this Inequality for Riemann-Stieltjes integral the following question came up. Suppose functions $f,g:[a,b] \to \mathbb{R}$ are such that $f$ is Riemann-Stieltjes integrable with respect to $g$. Suppose $g$ has bounded variation and $v_a^x(g)$ is the total variation on interval $[a,x]$.

Is it true that $f$ is integrable with respect to $v_a^x(g)$ (even if $f$ is not continuous)? If true how could this be proved?

The assertion is true.

To simplify notation let $$h(x) = v_a^x(g)$$. Given a partition $$P = (x_0,x_1,\ldots, x_n)$$ of $$[a,b]$$ consider the upper and lower sums,

$$U(P,f,h) = \sum_{j=1}^n M_j \, [h(x_j) - h(x_{j-1})], \\ L(P,f,h) = \sum_{j=1}^n m_j \, [h(x_j) - h(x_{j-1})],$$

where $$M_j = \sup_{x \in [x_{j-1},x_j]} f(x)$$ and $$m_j = \inf_{x \in [x_{j-1},x_j]} f(x).$$

To prove that the integral $$\int_a^b f \, dh$$ exists we can show that the difference $$U(P,f,h) - L(P,f,h)$$ can be made arbitrarily close to $$0$$ with a suitable choice for $$P$$.

Since $$f$$ is Riemann-Stieltjes integable with respect to $$g$$ it is bounded and $$|f(x)| \leqslant M$$ for all $$x \in [a,b].$$ Since $$g$$ has bounded variation, for any $$\epsilon > 0$$ there exists a partition $$P_\epsilon$$ such that for any refining partition $$P$$ we have

$$v_a^b(g) - \sum_{j=1}^n |g(x_j) - g(x_{j-1})| \leqslant \frac{\epsilon}{M}.$$

Since $$h(x_j) - h(x_{j-1}) = v_{x_{j-1}}^{x_j}(g) \geqslant |g(x_j) - g(x_{j-1})|$$, we have

$$U(P,f,h) - L(P,f,h) - \sum_{j=1}^n(M_j - m_j) |g(x_j) - g(x_{j-1})|\\ = \sum_{j=1}^n(M_j - m_j) [\,h(x_j) - h(x_{j-1}) - |g(x_j) - g(x_{j-1})|\,] \\ \leqslant 2M\sum_{j=1}^n [\,h(x_j) - h(x_{j-1}) - |g(x_j) - g(x_{j-1})|\,] \\ = 2M \left[ v_a^b(g) -\sum_{j=1}^n |g(x_j) - g(x_{j-1})|\,\right] \\ \leqslant 2 \epsilon.$$

Thus,

$$U(P,f,h) - L(P,f,h) < 2 \epsilon + \sum_{j=1}^n(M_j - m_j) |g(x_j) - g(x_{j-1})|.$$

If we can argue that the integrability of $$f$$ with respect to $$g$$ implies that sum on the RHS can be made smaller than $$\epsilon$$ for a suitable choice for $$P$$ (that refines $$P_\epsilon$$), then we are finished.

Unfortunately, this sum is not a difference of upper and lower sums due to the presence of the absolute value and the fact that, in general, $$g$$ may not be nonnegative.

Nevertheless, it is true that if $$P$$ is sufficiently fine, then

$$\sum_{j=1}^n (M_j - m_j)|g(x_j) - g(x_{j-1})| < \epsilon.$$

This can be shown by separately considering subintervals where $$g(x_j) - g(x_{j-1}) \geqslant 0$$ and where $$g(x_j) - g(x_{j-1}) < 0$$, and finding points $$\xi_j, \xi_j' \in [x_{j-1},x_j]$$ where, respectively, $$M_j - m_j < f(\xi_j) - f(\xi_j') + \epsilon/(2v_a^b(g))$$ and $$M_j - m_j < f(\xi_j') - f(\xi_j) + \epsilon/(2v_a^b(g)).$$

$$\sum_{j=1}^n (M_j - m_j)|g(x_j) - g(x_{j-1})| \\ \leqslant \sum_{j=1}^n [f(\xi_j) - f(\xi_j')] \, [g(x_j) - g(x_{j-1})] + \frac{\epsilon}{2 v_a^b(g)}\sum_{j=1}^n|g(x_j) - g(x_{j-1})| \\ \leqslant \sum_{j=1}^n [f(\xi_j) - f(\xi_j')] \, [g(x_j) - g(x_{j-1})] + \frac{\epsilon}{2}.$$

The first term on the RHS is a difference of Riemann-Stieltjes sums and is smaller that $$\epsilon/2$$ for $$P$$ sufficiently fine, completing the proof.

• Thanks. Can you explain further why the sum in the last inequality is smaller than $\epsilon/2$? – AlRacoon Jan 5 '18 at 17:43
• @AlRacoon: Note that $\sum_{j=1}^n [f(\xi_j) - f(\xi_j')] \, [g(x_j) - g(x_{j-1})] = S(P,f,g,T) - S(P,f,g,T')$ -- two different tagged sums corresponding to the same partition. So the absolute value of that sum is less than or equal to $|S(P,f,g,T) - \int_a^b f \, dg| + |S(P,f,g,T') - \int_a^b f \, dg| < \epsilon$, if $P$ is sufficiently fine since $f$ is integrable with respect to $g$. – RRL Jan 5 '18 at 20:18
• Nice and not-so obvious proof. +1 – Paramanand Singh Oct 14 '18 at 9:04
• How do you make sure that $M_j-m_j$ is $\leq$ some value where $g(x_j)-g(x_{j-1})<0$ (see another answer here). – Paramanand Singh Oct 14 '18 at 21:25
• @Matthieu: Thanks for spotting that typo. I will correct it. – RRL Mar 10 '19 at 2:04