# Step functions and Riemann-Stieltjes Integration

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.

• thanks for the edits! – 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

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

• 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 $1-1$. – RRL Feb 12 at 22:52
• Fantastic Thanks! – Matt Feb 12 at 23:03
• @MatteoLepur: You're welcome. – RRL Feb 12 at 23:10