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?

Solution 1:

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

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.