# Thursday August 22nd

> Todo: Find notes for first 15 minutes.

## Intervals Are Not Small

**Facts:**

- Countable $\implies$ Cantor, all intervals are not countable

- Meager $\implies$ Baire, all intervals are not meager

- Null $\implies$ Borel, all intervals are not null.

*Exercise:* 
Verify that $f$ is continuous at $x$ iff $\lim f(x_{n}) = f(x)$ for every sequence $\theset{x_{n}} \to x$.

## Discontinuities

**Definition:**
If $f: X \to \RR$, the *oscillation* of $f$ at $x \in X$ is given as

\begin{align*}
\omega_{f}(x) = \lim_{\delta \to \infty} \sup_{y \in B_{\delta}(x)} \abs{f(y) - f(z)}
.\end{align*}

*Exercise:* 
Show that $f$ is continuous at $x \iff \omega_f(x) = 0$.

We can then define points of discontinuity as
$$
D_f = \theset{x \in X \suchthat \omega_f(x) > 0} = \union_{n=1}^\infty\theset{x\in X \suchthat \omega_f(x) \geq \frac 1 n}
$$

*Exercise:* 
Show that $D_f$ is closed.

**Theorem 1:**
$f$ is monotone $\implies D_f$ is countable.

> Hint: we can't cover $\RR$ by uncountable many disjoint intervals.

**Theorem 2:** 
$D_f$ is always an $F_\sigma$ set.

> $\RR - \QQ$ is not at $F_\sigma$ set, i.e. one can not construct a function that is discontinuous on exactly this set.

**Theorem 3:**
$f$ is "1st class" $\implies D_f$ is meager.

> $f$ is **first class** if $f(x) = \lim_{n\to\infty} f_n(x)$ pointwise and each $f_n$ is continuous.

**Theorem 4 (Lebesgue Criterion):**
Let $f: [a, b] \to \RR$ be bounded, then $f$ is Riemann integrable iff $D_f$ is null.

> So the Dirichlet function is not Riemann integrable.

*Proof of theorems 1 and 2:*
**Exercise**.

*Proof of Theorem 3*

We want to show that $D_f$ is meager. 
We know it's some countable union of some sets, and it suffices to show that they are nowhere dense.

So let $F_n = \theset{x \suchthat \omega_f(x) \geq 0}$ for some fixed $n$. 
Let $I$ be an arbitrary closed interval, we will show that there exists a subinterval $J \subseteq I$ with $J \subseteq F_n^c$.

Consider
$$
E_k = \intersect_{i, j \leq k} \theset{x \suchthat \abs{f_i(x) - f_j(x)} \leq \frac 1 {5n}}
$$

> Motivation: this comes from working backwards from 4-5 triangle inequalities that will appear later.

Some observations: 
$E_k$ is closed by the continuity of the $f_i$ **(good exercise)**. 

We also have $E_k \subseteq E_{k+1}$. 
Moreover, $\union_k E_k = \RR$ because the $f_i \to f$ are Cauchy.

We'll now look for an interval entirely contained in the complement. 
Let $I \subset \RR$ be an interval, then write 
$$
I = \union_k (I\intersect E_k)
.$$ 

Baire tells us that $I$ is not meager, so at least one term appearing in this union is *not* nowhere dense, i.e. there is some $k$ for which $I \intersect E_k$ is not nowhere dense, i.e. it contains an open interval (it has a nonempty interior, and its already closed, and thus it contains an interval).

So let $J$ be this open interval. 
We want to show that $J \subseteq F_n^c$. 
If $x\in J$, then $x\in E_k$ as well, and so 
$$
\abs{f_i(x) - f_j(x)} \leq 1/5n \text{ for all } i,j \geq k
.$$ 

So let $i\to \infty$, so 
$$
\abs{f(x) - f_j(x)} \leq 1/5n \text{ for all } j\geq k
.$$

Now for any $x\in J$, there exists some interval $I(x) \subseteq J$ depending on $x$ such that $\abs{f(y) - f_k(x)} \leq 2/5n$. 

Now rewrite this as
$$
\abs{f(x) - f_j(x)} = \abs{f(y) - f_k(y) + f_k(y) - f_k(x)}
.$$

This implies that $\omega_f(x) \leq 4/5n$. 
$\qed$

## Integrability

*Proof of Theorem 4*:

Suppose that $f: [a,b] \to \RR$ is bounded. 

Recall that $f$ is Riemann integrable iff 
\begin{align*}
\forall ~\varepsilon \exists \text{ a partition } P_\varepsilon 
= \theset{a=x_1 \leq x_2 \leq \cdots x_n = b} &\text{ of } [a,b] \\
\text{ such that }U(f, P_\varepsilon) - L(f, \varepsilon) &\leq \varepsilon
,\end{align*}

where
$$
U(f, P_\varepsilon) - L(f, \varepsilon) \definedas
\sum_n \sup_{y, z \in [x_n, x_{n+1}]} \abs{f(y) - f(z)} (x_{n+1} - x_{n})
$$

$(\Rightarrow)$: 
Let $\varepsilon > 0$ and $n$ be fixed, and produce a partition $P_\varepsilon$ so that this sum is less than $\varepsilon / n$.

> Recall that we want to show that $F_n$ is null.

Now exclude from this sum all intervals that miss $F_n$, making it no bigger. 
We also know that in $F_n$, the sups are no greater than $1/n$,
$$
\varepsilon/n \geq \sum \text{stuff} \geq \sum \frac 1 n (x_{n+1} - x_n)
$$
$\qed$

$(\Leftarrow)$: 
Suppose $D_f$ is null and let $\varepsilon>0$ be arbitrary, we want to construct $P_\varepsilon$. 
Choose $n > 1/\varepsilon$ and $F_n \subseteq D_f$ is closed and bounded and thus compact. 

But a compact measure zero interval can in fact be covered by *finitely* many open intervals. 

So $F_n$ is covered by finitely many intervals $\theset{ I_n }_{n=1}^N$ such that $\sum \abs{I_n} \leq \varepsilon$.

Now if $x\not\in F_n$, then $\exists \delta(x) > 0$ where 
$$
\sup_{y,z \in B_\delta(x)} \abs{f(y) - f(z)} < \frac 1 n < \varepsilon.
$$

Since $(\union_j I_j)^c$ is compact, there's a finite cover $I_{N+1}, \cdots I_{N'}$ covering $F_n^c$.

$\qed$