Now we have that $\beta$ is continuous on [a,b] and of bounded variation. Now we have another function $g \in R(\beta)$ i.e it is Riemann-Stieljes integrable with respect to $\beta$.

Now we have $\gamma(x)$ = $\int_a^x gd\beta$ for $x\in[a,b]$. Now if $f$ is a continuous function we have to show that: $$\int_a^b fd\gamma = \int_a^b fgd\beta$$

My work so far:

For existence: So far I have shown that $\gamma$ is continuous on [a,b] and of bounded variation. There is a theorem in rudin that states that if your integrator is continuous and f is bounded then $f\in R(\gamma)$ but it also assumes $\gamma$ is monotonically increasing which is not necessarily the case here. Any help or hints would be appreciated

To show that $\int_a^b fd\gamma = \int_a^b fgd\beta$ :

It is enough to show that $\int_a^b fd\gamma = \int_a^b fgd\beta$ in the case where $\beta$ is monotonically increasing ( this was a hint given to me i don't really know why it is enough to just consider it when it is monotonically increasing)

Now $f \in R(\beta)$ since it is continuous (theorem in rudin). Now I have shown that the riemann sum $$S(f,\gamma,P) = \sum_{j=1}^n f(t_j)[(\gamma(x_j)-\gamma(x_{j-1})]$$ = $$\sum_{j=1}^n [f(t_j)(\int_{x_j-1}^{x_j} gd\beta)]$$ for some $t_j \in [x_{j-1}, x_j]$ and P is any parition.

Now i know that $fg$ are both Riemann Stieljes integrable with respect to $\alpha$ by thoerem 6.13 in Rudin. Hence for all $\epsilon > 0$ there exists a partition $P_\epsilon$ such that for any partition P finer than $P_\epsilon$ we have that:

$$ \sum_{j=1}^n f(t_j)g(t_j)[(\beta(x_j) - \beta(x_{j-1})] - \int_a^b fgd\beta < \epsilon $$

Now I can't seem to combine these two facts to get the desired result, any help or hints would be very appreciated thank you!

  • $\begingroup$ Please do not vandalize your own question. Once a question has been asked (and answered) it becomes part of our repository of knowledge, and from the very beginning, questions do no serve the sole purpose of the OP. $\endgroup$ Feb 26, 2018 at 15:14
  • $\begingroup$ sorry I am really high and drunk. I just wanted to delete this account and didn't know how $\endgroup$
    – user534253
    Feb 28, 2018 at 5:57
  • $\begingroup$ @user534253 What you did here is wrong. I'm flagging this. $\endgroup$
    – Deepak
    Feb 28, 2018 at 6:05

1 Answer 1


Any function of bounded variation is the difference of two monotonically increasing functions, so it is enough to prove this in the case that $\beta$ is monotonically increasing.

To finish, note that

$$\left| S(f,\gamma,P) - \int_a^b fg \, d\beta \right| \\= \left|\sum_{j=1}^n f(t_j)\int_{x_j-1}^{x_j} g \,d\beta - \sum_{j=1}^n\int_{x_j-1}^{x_j} f(y)g(y)\, \,d\beta \right| \\ \leqslant \sum_{j=1}^n \int_{x_j-1}^{x_j}|f(t_j)- f(y)| |g(y)| \,d\beta .$$

Since $g \in R(\beta)$ is bounded there exists $M > 0$ such that, $|g(y)| \leqslant M $ for all $y \in [a,b]$.

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

$$\left| S(f,\gamma,P) - \int_a^b fg \, d\beta \right| \leqslant M \sum_{j=1}^n \int_{x_j-1}^{x_j}|f(t_j)- f(y)| \,d\beta \\ \leqslant M \sum_{j=1}^n (M_j(f) - m_j(f))\int_{x_j-1}^{x_j} \,d\beta \\ = M \sum_{j=1}^n [\,M_j(f) - m_j(f)\,]\, [\, \beta(x_j)-\beta(x_{j-1}) \, ] \\ = M\left(U(f,\beta,P) - L(f,\beta,P) \right).$$

Given the hypotheses, we also have $f \in R(\beta)$. Thus, for any $\epsilon >0$ there is a partition $P_\epsilon$ such that if $P$ is any refinement, then $U(f,\beta,P) - L(f,\beta,P) < \epsilon/M$ and

$$\left| S(f,\gamma,P) - \int_a^b fg \, d\beta \right| < \epsilon,$$


$$\int_a^b f \, d\gamma = \int_a^b fg \, d\beta $$

  • $\begingroup$ Linearity is part of it. If $g$ is integrable with respect to $\beta_1$ and $\beta_2$, then it is integrable with respect to $\beta_1 - \beta_2$ and $\int g \, d(\beta_1 - \beta_2) = \int g \, d\beta_1 - \int g \, d\beta_2.$ If $\beta \in BV$ then $\beta = \beta_1 - \beta_2$ where $\beta_1,\beta_2$ are increasing. There are many such decompositions, but $g$ may not be integrable with respect to all $\beta_1, \beta_2$ individually. However it always will be true for one case: $\beta_1(x) = V_a^x(\beta)$, the total variation and $\beta_2 = \beta_1 - \beta$. $\endgroup$
    – RRL
    Feb 26, 2018 at 2:53
  • $\begingroup$ For the second question - yes. $\int f\, d(c_1 \alpha_1) = c_1\int f \, d\alpha_1$. $\endgroup$
    – RRL
    Feb 26, 2018 at 2:56
  • $\begingroup$ @user534253: You're welcome. Apostol is a more helpful book for understanding R-S integration including the results I discussed in the comment. $\endgroup$
    – RRL
    Feb 26, 2018 at 3:35
  • $\begingroup$ @user534253: All we need is that $f,g \in R(\beta)$ and $\beta \in BV([a,b])$. $\endgroup$
    – RRL
    Feb 28, 2018 at 3:28

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