I am reading about Riemann-Stieltjes Integrals in Carother's Real Analysis. I am trying to prove the following Theorem:

Theorem: $14.9$

Let $\alpha$ be continuous and increasing. Given $f$ $\in$ $R_{\alpha}([a,b])$ and $\epsilon >0 $.

There exists a step function $h$ on $[a,b]$ with $||h||_{\infty}$ $\leq$ $||f||_{\infty}$ such that $\int_{a}^{b}|f - h| d\alpha$ $<$ $\epsilon$.

Carother's uses the Riemann conditions to obtain a partition $P$ such that $U(f:P) - L(f:P)$ $<$ $\epsilon$. Using this partition, Carother's selects $t_i$ $\in$ $[x_{i-1}, x_i)$ and define the step function $h(x) = f(t_{i})$ for $i = 1,2,3,...n$. It follows clearly that $||h||_{\infty}$ $\leq$ $\|f||_{\infty}$ however, he makes the following statement that confuses me and concludes the proof using the statement:

"Since $\alpha$ is continuous, we have h $\in$ $R_{\alpha}([a,b])$"

I tried showing this myself by showing $U(h:P) - L(h:p) = \sum_{i=1}^{n}(M_{i} - m_{i})\Delta\alpha_{i} < \epsilon$ where $M_{i}$ and $m_{i}$ are the sup and inf of interval $i$ respectively. However, if $h(x)$ is a step function we have $M_{i} = m_{i}$ which implies $\sum_{i=1}^{n}(M_{i} - m_{i})\Delta\alpha_{i} = 0$. Therefore, I would not even consider the continuity of $\alpha$. Evidently, this is incorrect so if anyone could help explain it would be greatly appreciated.

  • $\begingroup$ thanks for the edits! $\endgroup$ – Matt Feb 12 at 23:05

The step function is $h = \sum_{j=1}^n f(t_j) \phi_j$ where

$$\phi_j(x) = \begin{cases}1,&x_{j=1} \leqslant x < x_j\\0, & \text{otherwise} \end{cases}$$

By linearity of integration it is enough to show that every function $\phi_j$ is integrable.

For any partition $P = (a=y_0,y_1, \ldots,y_{m-1}, y_m = b)$, since $\alpha $ is increasing , we have for some $p,q$,

$$\tag{*}0 \leqslant U(P,\phi_j,\alpha) - L(P,\phi_j, \alpha) \leqslant (\,\alpha(y_p)-\alpha(y_{p-1})\,) + (\,\alpha(y_q)-\alpha(y_{q-1})\,) $$

where the RHS bound is attained when $x_{j-1} \in (y_{p-1},y_p]$ and $x_j \in (y_{q-1},y_q)$.

Since $\alpha$ is continuous on the compact interval $[a,b]$ it is uniformly continuous and there exists $\delta > 0$ such that $|\alpha(u) - \alpha(v)| < \epsilon /2 $ when $|u-v| < \delta$.

Choosing the partition $P$ with norm $\|P\| < \delta$ we satisfy the integrability condition

$$U(P,\phi_j,\alpha) - L(P,\phi_j, \alpha) < \epsilon$$

  • $\begingroup$ The biggest that upper-lower sum difference can be occurs when the endpoints $x_{j-1}$ and $x_j$ are both placed in partition intervals where $\sup \phi_j - \inf \phi_j = 1 - 0$. For other partition intervals we get $0-0$ or $1-1$. $\endgroup$ – RRL Feb 12 at 22:52
  • $\begingroup$ Fantastic Thanks! $\endgroup$ – Matt Feb 12 at 23:03
  • $\begingroup$ @MatteoLepur: You're welcome. $\endgroup$ – RRL Feb 12 at 23:10

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.