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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.

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

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

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  • $\begingroup$ 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

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