Math4121 Lecture 8

Continue on Riemann-Stieltjes Integral

Integrable Functions

Theorem 6.9

If f is monotonic (increasing) on [a, b] and \alpha is continuous on [a, b], then f\in \mathscr{R}(\alpha) on [a, b].

Proof:

Given a partition P = \{a = x_0, x_1, \cdots, x_n = b\}, we have


M_i = \sup_{x\in [x_{i-i}, x_i]} f(x)\leq f(x_{i})


m_i = \inf_{x\in [x_{i-1}, x_i]} f(x)\geq f(x_{i-1})

So,


\begin{aligned}
U(P,f,\alpha) - L(P,f,\alpha) &= \sum_{i=1}^{n} (M_i - m_i)\Delta \alpha_i \\
&\leq \sum_{i=1}^{n} \left[ f(x_i) - f(x_{i-1}) \right] \left[ \alpha(x_i) - \alpha(x_{i-1}) \right] \\

&\leq \sum_{i=1}^{n} \left[ f(x_i) - f(x_{i-1}) \right](\sup_{x\in [x_{i-1}, x_i]} \alpha(x) - \inf_{x\in [x_{i-1}, x_i]} \alpha(x)) \\
&=U(P,\alpha,f) - L(P,\alpha,f)
\end{aligned}

By Theorem 6.8, \alpha\in \mathscr{R}(f), so for any \epsilon > 0, there exists a partition P such that


U(P,\alpha,f) - L(P,\alpha,f) < \epsilon

Therefore, U(P,f,\alpha) - L(P,f,\alpha)<U(P,\alpha,f) - L(P,\alpha,f) < \epsilon, so f\in \mathscr{R}(\alpha) on [a, b].

QED

Theorem 6.10

Suppose f is bounded on [a, b] and has finitely many points \{y_1, y_2, \cdots, y_J\} of discontinuity, and \alpha is continuous on [a, b]. Then f\in \mathscr{R}(\alpha) on [a, b].

Proof:

Since f is bounded, there exists a M>0 such that |f(x)|\leq M for all x\in [a, b].

Let \epsilon > 0. Since \alpha is continuous on [a, b], we can find some intervals [u_j,v_j]\subset (a,b) and y_j\in [u_j,v_j] and |\alpha(u_j) - \alpha(v_j)| < \epsilon for all j=1,2,\cdots,J.

Set K=[a,b]\setminus \bigcup_{j=1}^{J}(u_j,v_j). Since K is compact, f is uniformly continuous on K. Hence, there exists a \delta > 0 such that for any s,t\in K and |s-t|<\delta, we have |f(s)-f(t)|<\epsilon.

Let P=\{x_0,x_1,\cdots,x_n=b\} containing all the points u_j,v_j,\forall j=1,2,\cdots,J and \Delta x_i<\delta,\forall x_i\notin \{u_j,v_j,\forall j=1,2,\cdots,J\}.

Then,

If x_i=u_j for some j=1,2,\cdots,J, then M_i-m_i\leq M:=2\sup|f_x|. But \Delta \alpha_i\leq \epsilon for all i=1,2,\cdots,n.

If x_i\neq u_j for all j=1,2,\cdots,J, then by uniform continuity of f on K, we have M_i-m_i\leq \epsilon.

In either case, we have


\begin{aligned}
U(P,f,\alpha) - L(P,f,\alpha) &= \sum_{i=1}^{n} (M_i - m_i)\Delta \alpha_i \\
&\leq J M\epsilon + \sum_{i=1}^{n} \epsilon \Delta \alpha_i \\
&= \epsilon(J M + \sum_{i=1}^{n} \Delta \alpha_i)
\end{aligned}

Since \epsilon is arbitrary, we have U(P,f,\alpha) - L(P,f,\alpha) < \epsilon.

Therefore, f\in \mathscr{R}(\alpha) on [a, b].

QED

Theorem 6.11

Suppose f\in \mathscr{R}(\alpha) on [a, b], m\leq f(x)\leq M for all x\in [a, b], and \phi is continuous on [m, M], and let h(x)=\phi(f(x)) on [a, b]. Then h\in \mathscr{R}(\alpha) on [a, b].

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