# A Riemann integrable function must have infinitely many points of continuity

I was wondering whether anyone would be so kind as to briefly check my proof? I am supposed to prove the statement without using any theorems which would render the proof trivial.

If $\displaystyle\int_a^b f$ exists then $f$ has infinitely many points of continuity in $[a,b]$

For the sake of contradiction suppose $f$ has finitely many. It suffices to show that there is an interval $[u,v]\subset [a,b]$ over which $f$ is not Riemann integrable. Pick any $[u,b]\subset [a,b]$ such that $f$ is discontinuous everywhere in $[u,v]$. For any finite partition $D=\{x_1\cdots x_n\}$ of $[u,v]$ let:

$$s(f,D)=\sum_i (x_{i+1}-x_i)\inf_{x\in[x_{i},x_{i+1}]}f\quad\text{and}\quad S(f,D)=\sum_i (x_{i+1}-x_i)\sup_{x\in[x_{i},x_{i+1}]}f$$ We prove $\sup_D s(f,D) <\inf_D S(f,D)\;(1)$. To achieve this consider an arbitrary chain: $$D_0\subset D_1\subset \cdots$$ Pick any $I_0\subset I_1\subset \cdots$ where $I_k=[u_k,v_k]$ and $I_k$ is an interval of the partition $D_k$. Without loss of generality $\bigcap_i I_i=z$. Given that $f$ is discontinuous everywhere, there exists $\epsilon>0$ such that for any $\delta>0$ there exists $y$ such that $|z-y|<\delta$ yet $|f(z)-f(y)|\geq \epsilon$.

This immediately implies $\displaystyle\sup_{I_n} f>\inf_{I_n} f$ and hence $\displaystyle\sup_{D_n} s(f,D)<\inf_{D_n} S(f,D)$.

This is valid for any sequence $D_n$ so claim $(1)$ holds, contradicting the integrability of $f$.

• I wouldn't write "$|z-y|<\delta \implies$" rather, "$|z-y|<\delta$, yet...". Also "This immediately implies..." (Why?)
– Pedro
May 1, 2014 at 0:27
• @PedroTamaroff Thanks, adjusted. Is the argument reasonable otherwise? May 1, 2014 at 0:29
• You want to take $I_0\supset I_1\supset\cdots$; too. You wrote without loss of generality $\bigcap I_i=\color{red}{\{}z\color{red}{\}}$. This seems unnecessary, it suffices there is one there.
– Pedro
May 1, 2014 at 0:52
• A function is Riemann integrable iff it's set of discontinuity points has zero measure. May 1, 2014 at 2:50

One can also give a direct proof. Suppose $f:[0,1]\to \Bbb R$ is Riemann integrable.

Lemma Suppose that for a partition $P$; $\displaystyle D(f,P)=\sum_{k=0}^n (M_k-m_k)\Delta x_k<\varepsilon$. Then $M_k-m_k<\varepsilon$ for some $k$.

Proof Else $\displaystyle\varepsilon =\sum_{k=0}^n \varepsilon\Delta x_k\geqslant \sum_{k=0}^n (M_k-m_k)\Delta x_k$, proving the contrapositive.

Obs More generally, we get $M_k-m_k<(b-a)\varepsilon$ if the interval of integration is $[a,b]$.

We proceed to the proof of your claim.

Notation For a set $S$, we denote $o(f,S)=\sup\limits_{t,t'\in S}|f(t)-f(t')|$.

With this lemma at hand, we may start rolling: take $\varepsilon =1$. We know there exists a partition $P_1=\{0=x_0,x_1,\ldots,x_n=1\}$ with $D(f,P_1)<1$, since $f$ is Riemann integrable. By the lemma, there is a subinterval $I_i=[x_{i-1},x_i]$ where $o(f,I_i)<1$. If $i\neq 1$ or $n$, let $K_1=I_i$. Else, pick $x_0<x_0'<x_1$ or $x_{n-1}<x_n'<x_n$, and replace this in the endpoints of $K_i$. Now $f$ is integrable over $K_1$, so there exists a partition $P_2$ for which the sum in $K_1$, $D(f,P_2)<\frac 1 2$, which gives some subinterval $K_2$ where $o(f,K_2)<\dfrac 1 2|K_1|<\dfrac 1 2$, since $|K_1|<1$. Again, we replace endpoints if necessary. Since we're shrinking the domain, in any case the oscillation descreases, so we're safe. Inductively, we can construct a sequence of strictly nested closed bouned intervals $$K_1\supsetneq K_2\supsetneq K_2\supsetneq\cdots$$ where $o(f,K_i)<\frac 1 i$. By Cantor, there is $x_0\in\bigcap\limits_{j\geqslant 1} K_j$. I claim $f$ is continuous at $x_0$. (Proof?)

Now consider $[0,x_0],[x_0,1]$. Note that by construction $x_0\neq 0,1$, so repeating this on $[0,x_0]$ we obtain $x_1\neq x_0$ where $f$ is continuous, and so on. Using this, we may also show

Claim The set where $f$ is continuous is dense in $[0,1]$.

• For the interval $[a,b]$, don't we get $M_k-m_k<\frac{\epsilon}{(b-a)}$ in the lemma? Sep 8, 2016 at 9:52