Real Analysis integrability proof

Question

I need some tips for proving this proposition:

Let $f:[a,b] \to \mathbb{R}$ be a bounded function and let $t_k\in(a,b)$ be such that $t_k<t_{k+1}$ for $k=1,\dots,n-1$. Suppose that $f$ is continuous on $[a,b]\setminus{\{t_1,\dots,t_n\}}$. Prove that $f$ is integrable on $[a,b]$.

I thought that I could use Riemann's Criterion which is given as $\forall\epsilon>0$,

$$\begin{equation} U(f,P)-L(f,P)<\epsilon, \end{equation}$$

where $P$ is a partition of $[a,b]$ and $U(f,P)$ and $L(f,P)$ are the upper and lower sums of $f$ respectively.

Answer 1

Strategy: Cordon off the points where $f$ might be discontinous, and then apply the fact (which I'm guessing you know) that a continuous function is Riemann integrable to the restriction of $f$ to each of the subintervals left after the cordoning.

Out of laziness, I'll assume there's just one discontinuity point $t_1\in(a,b)$. Fix $\epsilon>0$. Because $f$ is bounded, there are numbers $m<M$ such that $m\le f(x)\le M$ for all $x\in[a,b]$. Now define $u=t_1-\delta$ and $v=t_1+\delta$, where $\delta>0$ is chosen to be smaller than $\epsilon/[6(M-m)]$ and also smaller than the distance between $t_1$ and either of $a$ or $b$. Now think of $[a,b]$ as broken up into three pieces: $[a,u], [u,v], [v,b]$. By hypothesis, $f$ is continuous on $[a,u]$ and on $[v,b]$. Therefore there is a partition $P_1$ of $[a,u]$ such that $U(f,P_1)-L(f,P_1)<\epsilon/3$ and a partition $P_2$ of $[v,b]$ such that $U(f,P_2)-L(f,P_3)<\epsilon/3$. Consider now the partition $P$ of $[a,b]$ consisting of $P_1\cup\{u,v\}\cup P_2$. Your task is to show that $U(f,P)-L(f,P)<\epsilon$.

Answer 2

This is how I completed the proof after John's suggestion:

$$\begin{equation} U(f,P)=U(f,P_1)+U(f,P_2)+\left(\sup_{t_1-\delta\leqslant x \leqslant t_1+\delta}f(x)\right)\left(t_1+\delta-(t_1-\delta)\right) \end{equation}$$ and $$\begin{equation} L(f,P)=L(f,P_1)+L(f,P_2)+\left(\inf_{t_1-\delta\leqslant x \leqslant t_1+\delta}f(x)\right)\left(t_1+\delta-(t_1-\delta)\right). \end{equation}$$

$\therefore$

$$\begin{equation} U(f,P)-L(f,P)\leqslant\frac{\epsilon}{3}+\frac{\epsilon}{3}+2\delta\left(\sup_{t_1-\delta\leqslant x \leqslant t_1+\delta}f(x)-\inf_{t_1-\delta\leqslant x \leqslant t_1+\delta}f(x)\right) \end{equation}$$

$$\begin{equation} \leqslant\frac{\epsilon}{3}+\frac{\epsilon}{3}+2\delta\left(M-m\right) \end{equation}$$

$$\begin{equation} <\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{2\epsilon(M-m)}{6(M-m)}=\epsilon \end{equation}$$

Answer 3

Since $f$ is bounded on $[a , b]$ there is a real number $M$ such that $|f(x)| \leq M$ for all $x \in [a, b]$.

Let $\varepsilon >0$ be given. Since the set $\{t_1 , t_2, \ldots, t_{N-1}, t_N \}$ has $N$ elements we cover it with $N$ disjoint intervals $\{(s_j, u_j)\}_{j=1}^N$ such that $(s_j, u_j) \subset [a,b]$ and $|s_j - u_j|< \frac{\varepsilon}{4MN}$ for $j=1, 2, \ldots, N$.

Now $K = [a, b] \setminus \bigcup_{j=1}^N (s_j, u_j)$ is compact and since $K$ contains no point of discontinuity we have that $f$ is uniformly continuous on $K$. Hence there exists $\delta>0$ so that $|f(x)-f(y)| < \frac{\varepsilon}{2(b-a)}$ whenever $x, y \in K$ and $|x-y|<\delta$.

Next we construct a partition $P=\{a, r_1, r_2, \ldots, r_{n-1}, b \}$ such that $s_j, u_j \in P$, $\, P \cap (s_j, u_j) = \emptyset$ and $\Delta r_i < \delta \,$ whenever $\, r_{i-1} \neq s_j$, for $j=1, 2, \ldots, N$.

Notice $\sup_{x \in I} \, f(x) - \inf_{x \in I} \, f(x) \leq 2M$ for any interval $I \subseteq [a, b]$. Hence\begin{equation} U(f,P)-L(f,P)<(b-a) \cdot \frac{\varepsilon}{2(b-a)} + 2MN \cdot \frac{\varepsilon}{4MN} = \varepsilon \end{equation}