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

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


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

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 $M_j - m_j < f(\xi_j') - f(\xi_j) + \epsilon/(2v_a^b(g)).$

This leads to

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

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.

  • $\begingroup$ Thanks. Can you explain further why the sum in the last inequality is smaller than $\epsilon/2$? $\endgroup$ – AlRacoon Jan 5 '18 at 17:43
  • $\begingroup$ @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 $g$. $\endgroup$ – RRL Jan 5 '18 at 20:18
  • $\begingroup$ Nice and not-so obvious proof. +1 $\endgroup$ – Paramanand Singh Oct 14 '18 at 9:04
  • $\begingroup$ 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). $\endgroup$ – Paramanand Singh Oct 14 '18 at 21:25
  • 1
    $\begingroup$ @Matthieu: Thanks for spotting that typo. I will correct it. $\endgroup$ – RRL Mar 10 '19 at 2:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.