Let $\;g:[a,b]\rightarrow \mathbb{C}\,$ be a function of bounded variation.

Let $\;f:[a,b]\rightarrow \mathbb{C}\,$ be a function Riemann-integrable along $g$.

Define $\,\alpha\left(x\right) = V_a^x \left(g\right)$, for all $\,x\in\left[a,b\right]$.

How do I prove that $\displaystyle\,\left\lvert\,\int f \,dg\, \right\rvert \leq \int \big\lvert\, f\, \big\rvert \,d\alpha$?

I have proven it for the case $\,g\,$ is monotonic, but in general how do I prove it?

Set $\,g=h+ik\,$ where $\,h,k\,$ are reals.

Then, set $\,h_1\left(x\right)=1/2\big(V_a^x\left(h\right) +h\left(x\right)\big)\, $ and $\,h_2\left(x\right)=1/2\big(V_a^x\left(h\right) - h\left(x\right)\big)\,$ and $\,k_1\left(x\right) = 1/2\big(V_a^x \left(k\right) +k\left(x\right)\big)\,$ and $\,k_2\left(x\right) = 1/2\big(V_a^x \left(k\right) - k\left(x\right)\big)$.

Then, \begin{align} \left\lvert\int_a^b f \,dg \right\rvert & = \left\lvert\int_a^b f \,dh_1 - \int_a^b f\, dh_2 + i\left(\int_a^b f\, dk_1 -\int_a^b f \,dk_2\right) \right\rvert \\ & \leq \left\lvert\int_a^b f \,dh_1 \right\rvert + \left\lvert\int_a^b f \,dh_2 \right\rvert + \left\lvert\int_a^b f\, dk_1 \right\rvert + \left\lvert\int_a^b f \,dk_2 \right\rvert \\ & \leq \int_a^b \big\lvert\, f\, \big\rvert \,dh_1 + \int_a^b \big\lvert\, f\, \big\rvert\,dh_2 + \int_a^b \big\lvert\, f\, \big\rvert\,dk_1 + \int_a^b \big\lvert\, f\, \big\rvert \,dk_2 \\ & = \int_a^b \big\lvert \,f\left(x\right) \big\rvert \,dV_a^x(h) + \int_a^b \big\lvert \,f\left(x\right)\big\rvert\,dV_a^x\left(k\right) \\ & = \int_a^b \big\lvert \,f\left(x\right)\big\rvert \, d\big( V_a^x\left(h\right) + V_a^x \left(k\right) \big). \end{align}

However, $\,V_a^x\left(h\right)+ V_a^x\left(k\right)\geq V_a^x\left(g\right)\,$ for all $\,x$, we cannot relate the above inequalities to prove the inequality in question.

How do I prove it? Thank you in advance.

  • $\begingroup$ It may be helpful to include the definition of $V_a^x$ in your question. Is it $\mu (\{t : x>g(t)>a\}$ or what? $\endgroup$ Sep 15, 2015 at 23:29
  • $\begingroup$ @PVAL $V_a^b(f)$ means the total variation of $f$ on $[a,b]$. I assume $f$ is Riemann-integrable, I don't mean Lebesgue measurable. $\endgroup$
    – Rubertos
    Sep 15, 2015 at 23:31
  • $\begingroup$ I'm sorry I haven't seen this stuff in a while. That is definitely something you shouldn't need to define. $\endgroup$ Sep 15, 2015 at 23:47

1 Answer 1



Let $g:[a,b]\rightarrow \mathbb{C}$ be of bounded variation and $f:[a,b]\rightarrow \mathbb{C}$ be Riemann-Stieltjes integrable along $g$.

Then, for every $\epsilon>0$, there exists a partition $P$ of $[a,b]$ such that every refinement $Q:=\{x_0,...,x_n\}$ of $P$ and $\mu\in \prod_{i=1}^n [x_{i-1},x_i]$, $|\int_a^b f dg - \sum_{i=1}^n f(\mu_i) (g(x_i)-g(x_{i-1}))|<\epsilon$.

The proof for lemma is simple. Start with a case [$f$:real, $g$:monotone] then [$f$:real, $g$:real bounded variation] then [$f$:complex, $g$:real bounded variation] them [$f$:complex, $g$:complex bounded variation].

To start a proof for the original question, fix $\epsilon$.

Then, there exists partitions such that $|\int f dg - \sum f(\mu_i) (g(x_i) -g(x_{i-1}))|<\epsilon$ and $|\int |f| dV_a^x(g) - \sum |f(\mu_i)| V_{x_{i-1}}^{x_i}(g) |<\epsilon$ respectively.

Now, take the common refinement of these partitions.

Then, $|\int f dg |< \epsilon + |\sum f(\mu_i) (g(x_i)- g(x_{i-1}))| \leq \epsilon + \sum |f(\mu_i)| |g(x_i) - g(x_{i-1})| \leq \epsilon + \sum f(\mu_i) V_{x_{i-1}}^{x_i}(g) < 2\epsilon + \int |f| dV_a^x (g)$.

This proves it.


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