# Function which is bounded and continuous except on a finite set of points is Riemann Integrable

I am trying to solve the following problem (Problem 7.2.15 of Bartle/Sherbert Book: Introduction to Real Analysis). The problem says:

If $f$ is a bounded function on $[a,b]$, and there is a finite set $E$ such that $f$ is continuous at every point of $[a,b]\E$, show that $f$ is Riemann integrable on $[a,b]$

My idea is to assumed that $E={c_{0},c_{1},...,c_{n}}$, and I am trying to show that the restriction of $f$ on each subinterval: $[c_{i},c_{i+1}]$ is Riemann integrable, and then use the additivity theorem to conclude that the function is Riemann integrable on $[a,b]$. However, I couldn't implement the previous idea. Any help is appreciated!

• That's the right approach. What can you say about the properties of $f$ on the interval $[c_i, c_{i+1}]$? (Or on the interval $[a, c_0]$ or $[c_n, b]$?) Does what you know about $f$ on those intervals tell you anything about its integrability? – John Hughes Jan 18 '14 at 21:33
• @John: The only thing I can say is that the function $f$ is bounded on $[c_{i},c_{i+1}]$ and continuous on $(c_{i},c_{i+1})$. (Hence, it is Riemann integrable on $(c_{i},c_{i+1})$) – user92040 Jan 18 '14 at 21:42
• @ user92040: The additivity theorem that I know says that if $f$ is integrable on $[a, b]$ with integral $A$ and on $[b, c]$ with integral $B$, then it's integrable on $[a, c]$ with integral $A + B$. With that, and some induction, you should be done. If your additivity theorem is weaker (e.g., requires integrability on $[a, c]$), then there's some real work left. – John Hughes Jan 18 '14 at 23:15
• @John: The issue here is that we can't say that the restriction of the function to the interval $[c_{i},c_{i+1}]$ is Riemann Integrable. We can Deduce that it is Riemann integrable on the open interval only (because the function is continuous on the open interval only) Can you show me how to prove that $f$ is Riemann Integrable on the closed interval? Thanks – user92040 Jan 18 '14 at 23:22
• Lemma: if $f$ is integrable on $[a, b]$ and $g = f$ on $(a, b)$, then $g$ is also integrable and has the same integral as $f$. Proof: Let $M = \max( |f(a) - g(a)|, |f(b) - g(b)|)$. Given $\epsilon > 0$ and a partition $P = \{t_0 = a, \ldots, t_n = b\}$, I'll construct a partition $P' = \{s_0, \ldots, s_{n+2}$ with $|U(f, P) - U(g, P')| < \epsilon\}$. A symmetric argument will apply to lower sums. To do so, I simply add the points $a + \frac{\epsilon}{2M}$ and $b - \frac{\epsilon}{2M}$ to the partition. Can you see how to show the difference of upper sums is small? – John Hughes Jan 18 '14 at 23:33

Because $f$ is bounded, it is enough to consider upper and lower sums for $f$ when determining whether or not $$\lim_{\|\mathscr{P}\|\rightarrow 0}\sum_{j}f(x_{j}^{\star})\Delta x_{j}$$ converges. The upper sum is defined in terms of $M_{j}=\sup_{x_{j-1}\le x_{j}^{\star}\le x_{j}} f(x_{j}^{\star})$ and the lower sum in terms of $m_{j}=\sup_{x_{j-1}\le x_{j}^{\star}\le x_{j}}f(x_{j}^{\star})$. Both $|m_{j}|$ and $|M_{j}|$ are bounded by $M$, where $M$ is a bound for $f$ on $[a,b]$. Clearly $|f(x)-f(y)| \le 2M$ for all $x,y\in [a,b]$.
The assumption is that the number $|E|$ of elements of $E$ is finite. Let $\delta > 0$, and form the union $E_{\delta}$ of all intervals $[e-\delta,e+\delta]$ for $e\in E$. The total length of these intervals is $2|E|\delta$. If $\|\mathscr{P}\| < \delta$, then the total contribution to the Riemann sum of intervals in $\mathscr{P}$ which contain at least one of point of $E_{\delta}$ is bounded by $4|E|\delta M$.
Now, let $\epsilon > 0$ be given. Choose $\delta > 0$ small enough that $8|E|\delta M < \epsilon/2$. Let $F_{\delta}$ be the closed set obtained by subtracting the intervals $(e-\delta,e+\delta)$ from $[a,b]$. This leaves a finite, disjoint set of closed intervals on which $f$ is continuous. So there exists $\delta' > 0$ such that $$|f(x)-f(y)| < \frac{\epsilon}{2(b-a)}$$ whenever $|x-y| < \delta'$ and whenever $x,y$ are in the same interval of $F_{\delta}$. If $\mathscr{P}$ is any partition of $[a,b]$ with $\|\mathscr{P}\| < \min\{\delta,\delta'\}$, then the difference of upper and lower sums over $\mathscr{P}$ is bounded by $$\frac{\epsilon}{2(b-a)}\left(\sum_{\{j \;:\; [x_{j-1},x_{j}]\cap E_{\delta}=\emptyset\}}\Delta x_{j}\right)+ 2M\left(\sum_{\{ j\; : \; [x_{j-1},x_{j}]\cap E_{\delta}\ne \emptyset\}}\Delta x_{j}\right)$$ $$\le \frac{\epsilon}{2(b-a)}(b-a)+2M(|E|2\delta+2|E|\|\mathscr{P}\|) \le \epsilon/2 + 2M|E|4\delta < \epsilon/2+\epsilon/2.$$ Because $\epsilon > 0$ was arbitrary, then it follows that $f$ is Riemann integrable on $[a,b]$.
• Thanks for your answer! Can you, please, explain why you have $(|E|2\delta+2|E|\|\mathscr{P}\|)$ instead of $|E|2\delta$. I don't understand why you included the term $2|E|\|\mathscr{P}\|$ in last inequality? – user92040 Jan 19 '14 at 1:10
• You said "the total contribution to the Riemann sum of intervals in 𝒫 which contain at least one of point of Eδ is bounded by $4|E|δM$". Shouldn't it be just $2|E|δM$? I appreciate if you explain that too. Thanks! – user92040 Jan 19 '14 at 1:12
• user92040: You're welcome. In the first part, I had stated, "if $\|\mathcal{P}\|<\delta$" to simplify the overview. I'm calling more attention to that in the last part. The second sum of "exception intervals" are intervals from the partition which hit $[e-\delta,e+\delta]$, the total length of which is bounded by $\|\mathscr{P}\|(\mbox{on left})+2\delta+\|\mathscr{P}\|(\mbox{on right})$; but $\|\mathscr{P}\| < \delta$, which makes everything okay. Just calling attention to this subtle point. This is need to keep uniform continuity of $f$ from being a moving target on the rest of $[a,b]$. – DisintegratingByParts Jan 19 '14 at 3:10
• user92040: This is a subtle point. You have to pick intervals $[e-\delta,e+\delta]$ to exclude first. Then you can get uniform continuity over the remaining part of $[a,b]$. So you cannot invoke uniform continuity over a larger region once $\delta$ is chosen, because $\delta'$ depends on $\delta$. That's a very tricky part of getting such a proof right. That's why I have to exclude more than enough from the first sum where I am invoking uniform continuity in order to bound the difference of upper and lower sums. – DisintegratingByParts Jan 19 '14 at 3:24
We prove the claim by strong induction on the number of discontinuities of $f$ on $[a, b]$. The base case $n = 1$ is that $f$ has only one discontinuity $c$ of $[a, b]$. So $c$ can be $a, b,$ or a middle point that is $a < c < b$. If $c$ is the middle point, then we can prove that $f$ is integrable on $[a, c]$, and then on $[c, b]$, then $f$ is integrable on $[a, b]$ by addition property of Riemann integrable function. So it suffices to prove for the case $c = a$ or $c = b$. Both cases are done similarly so we can simplify the argument by taking $c = a$. Now we can choose a point $x(1)$ of $[a, b]$ close to a such that $(M(1) - m(1))(x(1) - a) < \frac{e}{2}$ and this is possible because $M(1) - m(1) < 2K$ with $K = \sup \{ f(x) : x \in [a, b]\}$. Next on the interval $[x(1), b]$, $f$ is continuous since there is no discontinuity here so it is integrable here, and thus there is a partition $Q$ of $[x(1), b]$ such that $U(Q,f) - L(Q,f) < \frac{e}{2}$. Now let $P = Q \cup \{a\}$ be a partition of $[a, b]$, then $U(P,f) - L (P,f) = (M(1) - m(1))(x(1) - a) + U(Q,f) - L(Q,f) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$. So $f$ is integrable on $[a, b]$. So suppose that $f$ is integrable on $[a, b]$ with $n$ discontinuities. Now if $f$ has $n + 1$ points of discontinuities on $[a, b]$, then let say $x^*$ be one of these points, then consider breaking $[a, b]$ into $3$ sub-intervals $[a, x^*], [x^*, x^{**}], and [x^{**}, b]$ where $x^*$ and $x^{**}$ are two discontinuities points of $[a, b]$ and appeal the inductive step on each of these sub-intervals we get $f$ is integrable on each of these sub-intervals and therefore $f$ is integrable on $[a, b]$.