# Show that the Riemann Stieljes integral of fd$\gamma$ exists

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!

• 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. Feb 26, 2018 at 15:14
• sorry I am really high and drunk. I just wanted to delete this account and didn't know how
– user534253
Feb 28, 2018 at 5:57
• @user534253 What you did here is wrong. I'm flagging this. Feb 28, 2018 at 6:05

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

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

proving

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

• 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 $\beta_2 = \beta_1 - \beta$.
– RRL
Feb 26, 2018 at 2:53
• For the second question - yes. $\int f\, d(c_1 \alpha_1) = c_1\int f \, d\alpha_1$.
– RRL
Feb 26, 2018 at 2:56
• @user534253: You're welcome. Apostol is a more helpful book for understanding R-S integration including the results I discussed in the comment.
– RRL
Feb 26, 2018 at 3:35
• @user534253: All we need is that $f,g \in R(\beta)$ and $\beta \in BV([a,b])$.
– RRL
Feb 28, 2018 at 3:28