Rudin real anlysis: prove that $f$ is Riemann integrable

Question

Let $f : [0,1] →\Bbb R$ satisfy the property that for every $ε>0$ the set $\{x∈[0,1]: |f(x)|\ge ε\}$ is finite. Prove that for every continuous increasing function $α :[0,1]→\Bbb R$, that $f$ is Riemann integral and integral $\int_0^1f\,dα=0$.

I am confused about what is the meaning of the the set $\{x∈[0,1]: |f(x)|\ge ε\}$ is finite. And I want to apply theorem 6.10(rudin):suppose f is bounded on[a,b], f has only finitely many points of discontinuity on[a,b], and α is continuous at every point at which f is discontinuous. Then f is Riemann integral.

Answer 1

First of all, it's not true that $f$ has only finitely many points of discontinuity. For example, $g : [0,1] \to \mathbf R$ defined by

$$ g(x) = \begin{cases} 0 & x \notin \mathbf Q \\ \frac{1}{q} & x = \frac pq \text{ where } \gcd(p,q) = 1 \end{cases} $$

Notice that there are only finitely many rational numbers with denominator $\le N$ for any given integer $N$.

On the other hand, let

$$ A = \{x : f(x) \ne 0\} = \bigcup_{n \ge 1} \left\{x \in [0,1] : |f(x)| \ge \frac1n \right\} $$

You can show that $f$ is continuous on $A^c$. (Hint: it is a similar proof to $g$ being continuous at irrational numbers.)

For $\alpha(x) = x$, the theorem from the Lebesgue theory portion of Rudin says that $f$ is Riemann integrable if and only if it is continuous almost everywhere. But we don't need to use this theorem, we can prove it by hand.

Here's the sketch when $\alpha(x) = x$ and I will leave it to you to fill in the details and extend to a general continuous and increasing function $\alpha$.

There are finitely many points, say $x_1,\dots,x_N$ in $[0,1]$ where $|f(x)| > \delta$. Surround these points $x_1,\dots,x_N$ by small intervals, $I_1,\dots,I_N$. Let $J = [0,1] \setminus (I_1 \cup \cdots \cup I_N)$. Then a Riemann sum over this partition is bounded by $$ \left(\sup_J |f(x)|\right) \operatorname{length}(J) + \left(\sup_{I_1} |f(x)|\right) \operatorname{length}(I_1) + \dots + \left(\sup_{I_N} |f(x)|\right) \operatorname{length}(I_N) $$ If you play with the numbers right you can get this $< \varepsilon$.