After a lecture today, I just wanted to confirm that I understand the proof of the following:
If $f: [a,b] \to \mathbb{R}$ is bounded and continuous and has finitely many discontinuities, $f \in \mathscr{R}$.
Proof:
Let $\{x_1,\ldots x_l\}$ be the set of discontinuities of $f$. Let $\epsilon > 0$. Let $M$ be such that $|f| \leq M$. Let $\delta > 0$ be such that $2(k+2)M\delta < \frac{\epsilon}{2}$. Now, consider the interval $$B= [a,b] \setminus \left( (x_1-\delta, x_1 + \delta) \cup (x_2-\delta, x_2+\delta) \cup \cdots \cup (x_k-\delta, x_k+\delta) \right),$$ clearly $B$ is compact by Heine-Borel. Thus, $f|B :B \to \mathbb{R}$ is uniformly continuous. Thus, there exists $\delta > 0$ so that, if $|x-y| < \delta$, $|f(x) – f(y)| < \frac{\epsilon}{2(b-a)}$. Now, let $n$ be a positive integer such that $\frac{b-a}{n} < \delta$. Consider the partition
$$P:= \left\{a=x_0, a+\frac{x_1-\delta-a}{n}, a+2\frac{x_1-\delta}{n}, \ldots x_1- \delta, x_1+\delta, \ldots, x_k+\delta , x_k+\delta+\frac{b-(x_k+\delta)}{n}, \ldots, b \right\}$$ That is, we have two types of intervals. One type is of the form $[x_i – \delta, x_i+\delta]$ while we chop what is between $x_i+\delta$ and $x_{i+1} – \delta$ into $n$ parts. Now here's where I get a bit confused. I suppose we chop this middle piece up even further, although I'm not sure why. Consider the intervals
$$D_i = \left[x_i+\delta+m\left(\frac{x_{i+1} – x_i -2\delta}{n}\right), x_i+\delta+(m+1)\left(\frac{x_{i+1} – x_i -2\delta}{n}\right)\right]$$
Call $\frac{x_{i+1} – x_i -2\delta}{n} =T$. Then, if $y_1,y_2 \in D_k$, then $$|y_1-y_2| \leq T \leq \frac{b-a}{n} < \delta$$.
so that if $y_1,y_2 \in D_i$, the uniform continuity condition holds.
Now, for any $y \in D_i$, $f(y) = f(y) – f(y_1) + f(y_1)$ so that $f(y) \leq \frac{\epsilon}{2(b-a)} + f(y_1)$ for all $y \in D_i$. Thus, $$\sup_{x \in D_i} f(x) \leq \frac{\epsilon}{2(b-a)} + f(y_1)$$ and $$\inf_{x \in D_k} f(x) \geq -\frac{\epsilon}{2(b-a)} + f(y_1)$$. Thus, $$\sup_{x \in D_i} f(x)- \inf_{x \in D_k} f(x) \leq \frac{\epsilon}{(b-a)}.$$ Now, we need show that $U(f,P) – L(f,P) < \epsilon$. Let $M_i = \sup_{x \in \Delta_i} f(x)$ and $m_i$ similarly. We have that
$$ \begin{align}
U(f,P) – L(f,P) &= \sum_{P} (M_i-m_i)\Delta_i \\
&= \sum_{[x_i-\delta,x_i+\delta]} (M_i-m_i)\Delta_i + \sum_{D_i} (M_i-m_i)\Delta_i \\&\leq
\sum_{[x_i-\delta,x_i+\delta]} 4M\delta + \sum_{D_i} \frac{\epsilon}{(b-a)}\frac{b-a}{n} \\&=
4Mk\delta + (k+1)n\frac{\epsilon}{n} \\&< \epsilon
\end{align}$$
Thus, $f \in \mathscr{R}$.
Now, this proof seems a bit mathemagical to me. I understand that it works, just not why. I see that we construct little intervals around the discontinuities in order to "ignore" them, but I don't see, for example, why we need to create the $D_i$ intervals. Why do we need to chop the interval between the $x_i+\delta$ and $x_{i+1} -\delta$ into $n$ pieces. Why can't we just consider the middle and only the middle? Any insight to the intuition would be greatly appreciated.
Best Answer
It is perhaps easiest to consider the case where $f$ has exactly one discontinuity, say at $x_0 \in (a,b)$. Fix $\epsilon > 0, \delta > 0$ with $\delta$ to be chosen later.
Since $f\lvert_{[a,x_0]}$ is Riemann integrable, there is a partition $P_1$ of $[a,x_0-\delta]$ such that $$ U(f,P_1) - L(f,P_1) < \epsilon $$ Similarly, there is a partition $P_2$ of $[x_0+\delta,b]$ such that $$ U(f,P_2) - L(f,P_2) < \epsilon $$ Now one would hope that $P:= P_1\cup P_2\cup\{[x_0-\delta,x_0+\delta]\}$ would be a partition of $[a,b]$ which would satisfy $$ U(f,P) - L(f,P) < \epsilon \qquad (\ast) $$ Of course, the problem is, one needs to control the term contributed by $[x_0-\delta,x_0+\delta]$. Since $f$ is discontinuous at $x_0$, you cannot control the max and min on this interval, so you just control the width of the interval. ie. If $M > 0$ is such that $|f| \leq M$ on $[a,b]$, then you choose $\delta > 0$ such that $$ 4M\delta < \epsilon $$ Then, if $M_0$ and $m_0$ denote the $\sup$ and $\inf$ of $f$ over the interval $[x_0-\delta,x_0+\delta]$, then you have $$ (M_0 - m_0)(x_0+\delta - (x_0-\delta)) \leq 4M\delta < \epsilon $$ Now you should get $(\ast)$ (with perhaps a $3\epsilon$ instead of $\epsilon$).
Does this help?