Theorem 6.10 Suppose $f$ is bounded on $[a, b]$, $f$ has only finitely many points of discontinuity on $[a, b]$, and $\alpha$ is continuous at every point at which $f$ is discontinuous. Then $f \in \mathscr{R}(\alpha)$.
Proof: Let $\varepsilon > 0$ be given. Put $M = \sup \left\vert f(x) \right\vert$, let $E$ be the set of points at which $f$ is discontinuous. Since $E$ is finite and $\alpha$ is continuous at every point of $E$, we can cover $E$ by finitely many disjoint intervals $\left[ u_j, v_j \right] \subset [a, b]$ such that the sum of the corresponding differences $\alpha\left(v_j\right) – \alpha \left( u_j \right)$ is less than $\varepsilon$. Furthermore, we can place these intervals in such a way that every point of $E \cap (a, b)$ lies in the interior of some $\left[ u_j, v_j \right]$.
Remove the segments $\left( u_j, v_j \right)$ from $[a, b]$. The remaining set $K$ is compact. Hence $f$ is uniformly continuous on $K$, and there exists $\delta > 0$ such that $\left\vert f(s) – f(t) \right\vert < \varepsilon$ if $s \in K$, $t \in K$, $\left\vert s-t \right\vert < \delta$.
Now form a partition $P = \left\{ x_0, x_1, \ldots, x_n \right\}$ of $[a, b]$, as follows: Each $u_j$ occurs in $P$. Each $v_j$ occurs in $P$. No point of any segment $\left( u_j, v_j \right)$ occurs in $P$. If $x_{i-1}$ is not one of the $u_j$, then $\Delta x_i < \delta$.
Note that $M_i – m_i \leq 2M$ for every $i$, and that $M_i – m_i \leq \varepsilon$ unless $x_{i-1}$ is one of the $u_j$. Hence, as in the proof of Theorem 6.8,
$$ U(P, f, \alpha) – L(P, f, \alpha) \leq \left[ \alpha(b) – \alpha(a) \right] \varepsilon + 2M \varepsilon.$$
Since $\varepsilon$ is arbitrary, Theorem 6.6 shows that $f \in \mathscr{R}(\alpha)$.
I don’t “really” understand the proof. To be more specific
(1)Since $E$ is finite and $\alpha$ is continuous at every point of $E$, we can cover $E$ by finitely many disjoint intervals $\left[ u_j, v_j \right] \subset [a, b]$ such that the sum of the corresponding differences $\alpha\left(v_j\right) – \alpha \left( u_j \right)$ is less than $\varepsilon$.
(2)Furthermore, we can place these intervals in such a way that every point of $E \cap (a, b)$ lies in the interior of some $\left[ u_j, v_j \right]$.
(3)If $x_{i-1}$ is not one of the $u_j$, then $\Delta x_i < \delta$.
(4)Note that $M_i – m_i \leq 2M$ for every $i$.
(5)Hence, as in the proof of Theorem 6.8,
$ U(P, f, \alpha) – L(P, f, \alpha) \leq \left[ \alpha(b) – \alpha(a) \right] \varepsilon + 2M \varepsilon.$
Now I will try to address more explicitly what I’m not able to understand. I will sort of try to prove each sentences from above.
(1) let $\epsilon \gt 0$ given and $E=\{ d_1 ,d_2,….,d_{k}\}$; $k \in \Bbb{N}$. Since $\alpha$ is continuous at $d_{j}$, $\exists \delta_{j} \gt 0$ s.t $|x-d_j| \lt \delta_{j} \Rightarrow |\alpha(x)-\alpha(d_j)| \lt \epsilon$. $x-\delta_j \lt d_j \lt \delta_j +x$, choose $u_j \in (x-\delta_j, d_j)$ & $v_j \in (d_j, \delta_j +x)$ such that $v_{j-1} \lt u_{j}$ & $v_j \lt u_{j+1}$ respectively. Thus, $\alpha(v_j) – \alpha(u_j) \lt \epsilon$ and each $[u_j , v_j]$ contains $d_j$ and $[u_m , v_m] \cap [u_n , v_n] = \phi $, $m \neq n$. In other words, $[u_j , v_j]$ are disjoint. $\{[u_j , v_j]|j \in J_{k}\}$ obviously cover $E$.
(2) I don’t satisfactorily understand that sentence.
(3) Is it one of the axiom of construction of partition $P$ of $[a,b]$?
(4) That inequality seems obvious but I can’t prove it formally. I won’t be surprise if it is a few liner proof.
(5) Just the way inequality is written. Perhaps we are using the same “technique” as in theorem 6.11. Which is, to divide the members of partition $P$ into two category: $i \in A$ if $x_{i-1} \neq u_j, \forall j$ and $i \in B$ if $x_{i-1}=u_{j}$, for some $j$. $M_i – m_i \leq \epsilon$ for $i \in A$, since $\Delta x_i \lt \delta$(Proof: $|s_i – t_i|\leq \Delta x_i \lt \delta \Rightarrow |f(s_i)-f(t_i)| \lt \epsilon \Rightarrow M_i – m_i \leq \epsilon$. But I’m not sure about this proof because $s_i, t_i \in K$ ). So, $ \sum_{i \in A} (M_i – m_i) \Delta \alpha_i \leq \epsilon (\alpha(b) – \alpha(a))$. what’s about $\sum_{i \in B} \Delta \alpha_i$?
Best Answer
(1)(2)
Assume $E=\{t_1,...,t_n\}$. It is possible to choose $\delta'>0$ so that $$\lvert\alpha(t)-\alpha(t_i)\rvert<\frac{\epsilon}{2n}\quad(i=1,2,...,n)$$ provided that $\lvert t-t_i\rvert<\delta'$, $t\in[a,b]$. Hence if $\lvert t-t_i\rvert<\delta'$ and $\lvert s-t_i\rvert<\delta'$, $$\lvert \alpha(t)-\alpha(s)\rvert<\frac{\epsilon}{n}\mbox{.}$$ For $t\in E\cap(a,b)$, choose $u_j,v_j\in(a,b)$ with $t-\delta'<u_j<t<v_j<t+\delta'$. If $a\in E$, let $u_j=a$ and choose $v_j$ with $u_j=a<v_j<a+\delta'$. If $b\in E$, let $v_j=b$ and choose $u_j$ with $b-\delta'<u_j<b=v_j$. It follows that $$\sum_j\lvert\alpha(u_j)-\alpha(v_j)\rvert<n\frac{\epsilon}{n}= \epsilon \mbox{.}$$
(3)
$P$ is constructed in such a way that the uniform continuity of $f$ can be applied.
(4)
By definition of $M$, $$-M\leq f(x)\leq M\quad (a\leq x\leq b)\mbox{,}$$ so that $$-M\leq m_i\leq M_i\leq M\mbox{,}$$ and hence $M_i-m_i\leq 2M$.
(5)
The contribution of the continuous part of $f$ to $U-L$ can be estimated as follows $$\sum (M_i-m_i)\Delta\alpha_i\leq\epsilon\sum\Delta\alpha_i\leq\epsilon(\alpha(b)-\alpha(a))\mbox{,}$$ where the sum is taken over all $i$ such that $x_i$ is not one of the $u$'s. The other part can be estimated as follows $$\sum(M_i-m_i)\Delta\alpha_i\leq 2M\sum \Delta\alpha_i<2M\epsilon\mbox{.}$$