# Ngô Quốc Anh

## April 5, 2015

### The set of continuous points of Riemann integrable functions is dense

Filed under: Uncategorized — Ngô Quốc Anh @ 15:03

In this note, we prove that the set of continuous point of Riemann integrable functions $f$ on some interval $[a,b]$ is dense in $[a,b]$. Our proof start with the following simple observation.

Lemma: Assume that $P=\{t_0=a,...,t_n=b\}$ is a partition of $[a,b]$ such that

$\displaystyle U(f,P)-L(f,P)<\frac{b-a}m$

for some $m$; then there exists some index $i$ such that $M_i-m_i < \frac 1m$ where $M_i$ and $m_i$ are the supremum and infimum of $f$ over the subinterval $[t_{i-1},t_i]$.

We now prove this result.

Proof of Lemma: By contradiction, we would have $M_i-m_i \geqslant \frac 1m$ for all $i$; hence

$\displaystyle \frac{b-a}m = \sum_{i} \frac{t_i-t_{i-1}}{m}\leqslant \sum_{i} \big(M_i-m_i\big)(t_i-t_{i-1})=U(f,P)-L(f,P),$

We now state our main result:

Theorem. Let $f$ be Riemann integrable over $[a,b]$. Define

$\displaystyle \Gamma = \{ x\in [a,b] : f \text{ is continuous at } x\}$

Then $\Gamma$ is dense in $[a,b]$.

We now prove the theorem.

Proof of Theorem: Let $P$ be a partition of $[a,b]$ such that $U-L, where $U$ and $L$ are the upper and lower Darboux sums of $f$. By the lemma, there is some index $i$ for which $M_i - m_i<1$.

Step 1. Narrow $[a,b]$ to obtain a subinterval whereas $\sup f - \inf f$ is relatively small.

To see this, we first select

$\displaystyle I_1=(a_1,b_1)=\begin{cases}(t_{i-1}, t_i), & \text{ if } i \ne 1 \text{ or } i\ne n\\ (a_1, t_1), & \text{ if } i =1 \text{ and for some } a_1 \in (a,t_1) \\ (t_{n-1}, b_1), & \text{ if } i =n \text{ and for some } b_1 \in (t_{n-1}, b). \end{cases}$

If $I_1$ happens to be $(t_{i-1}, t_i)$, then we immediately get $\sup_{I_1} f -\inf_{I_1}f <1$. Otherwise, it will be the case that

$\displaystyle \sup_{A'} f \geqslant \sup_A f$

where $A'=[t_0,t_1]$ (resp. $[t_{n-1},t_n]$) and $A=[a_1,t_1]$ (resp. $[t_{n-1},b_1]$) and also that

$\displaystyle\inf_{B'} f \leqslant \inf_B f$

where $B'=[t_0,t_1]$ (resp. $[t_{n-1},t_n]$) and $B=[a_1,t_1]$ (resp. $[t_{n-1},b_1]$). Thus, in any case, there would exist $a and

$\displaystyle \sup_{I_1} f -\inf_{I_1}f <1$

where $I_1=[a_1,b_1]$. Since $f$ is integrable over $[a,b]$, it is integrable over $[a_1,b_1]$.

Step 2. Repeat the above argument to get a sequence of strictly nested subintervals.

The very same reasoning but with $m=2$ gives us $a_2$, $b_2$ with $a and with

$\displaystyle \sup_{I_2} f-\inf_{I_2}f< \frac 12$

Continuing the process gives us a sequence of nested closed intervals $\{I_n:n\in\mathbb N\}$ such that

$\displaystyle \sup_{I_n}f - \inf_{I_n} f< \frac 1n.$

By the Cantor theorem,

$\displaystyle \bigcap_{n \in \mathbb N} I_n \ne \emptyset.$

Step 3. Existence of points of continuity in every subinterval.

Let then $\mu \in \bigcap_n I_n$. Since the sequences of $\{b_i\}_i$ and $\{a_i\}_i$ are strictly monotonic, it follows that $\mu$ is none of the extrema. Let $\varepsilon>0$ be given, and choose $n\in \mathbb N$ such that $\frac 1n<\varepsilon$. Keep in mind that $\mu \in I_n = (a_n,b_n)$ for any $n$.

Let $\delta>0$ such that if $|\mu-x|<\delta$, then $x\in I_n$. Since $\inf f \leqslant f \leqslant \sup f$, for each $x\in I_n$, it follows that if $|x-\mu|<\delta$ then

$\displaystyle |f(x)-f(\mu)|<\frac 1n<\varepsilon,$

and thus $f$ is continuous at $x=\mu$. Since $f$ is integrable over $[a,b]$ it also is integrable over $[a,\mu]$ and $[\mu,b]$. Set $\mu=\mu_1$.

By the previous reasoning, we obtain two new different points of continuity, $\mu_2$ and $\mu_3$. Having repeated this $j$ times we would obtain $2^{j+1}-1$ continuity points. Thus $f$ cannot be continuous over a finite number of points $M$, for it would mean there exists $M$ such that $2^{j+1}-1 for each $j\in \mathbb N$. It follows $\Gamma$ is infinite.

Step 4. $\Gamma$ is dense in $[a,b]$.

Let $x\in [a,b]$ and consider the sequence

$\displaystyle\sigma_0=[a,x], \quad \sigma_n=\left[x-\frac{x+a}{2^n},x\right].$

The previous argument means there exists for each $n$ a $\gamma_n \in \sigma_n$ with $\gamma_n \in \Gamma$. But then given $\varepsilon>0$ we can take obtain an $N$ large enough such that $\frac{|x+a|}{2^n}<\varepsilon$ whenever $n\geqslant N$ and thus

$\displaystyle |x-\gamma_n| \leqslant \frac{|x+a|}{2^n}<\varepsilon$

for $n\geqslant N$, that is, $\gamma_n \to x$. Since $x$ was arbitrarily chosen, $\Gamma$ is dense over $[a,b]$. This completes the proof.

Source:

2. $\displaystyle \sigma_n = \left[x-\frac{x-a}{2^n},x\right]$? 🙂