# Tuesday August 27th

## Nowhere Density

Recall **Baire's Theorem**: 
$\RR$ can not be written as a countable union of nowhere dense sets. 

A subset $A\subseteq \RR$ is *nowhere dense* 
$\iff$  every interval contains a subinterval which lies entirely in $A^c$ 
$\iff$ $\overline A$ has empty interior
$\iff$ $\overline A$ contains no open intervals.

*Exercise:* 
Show that these definitions are equivalent.

**Corollary:** 
$\RR\setminus\QQ$ is not an $F_\sigma$ set. 


*Proof:*
Suppose it was, so 
$$
\RR\setminus \QQ = \union_{n\in\NN} F_n \text{ with } A_n \text{ closed }
.$$ 
Then 
$$
\RR = \left( \union_n F_n\right) \union \left( \union_i \theset{q_i} \right)
\text{ where } \QQ = \union \theset{q_i}
.$$

This exhibits $\RR$ as a countable union of closed sets. 
But the $F_n$ are nowhere dense, since if they contained in interval they'd also contain a rational.
$\qed$

*Exercise*: 
Show that $F_n$ are nowhere dense by constructing a sequence of elements in $F_n$ that converges to an element in $F_n^c \subset \QQ$.


## Riemann Integration

**Some nice properties:**

- Good for approximation (vertical strips)

- Many functions are in $\mathcal R$, e.g. continuous functions.

- $\mathcal{R} ([a, b])$ is a vector space

- The integral is an element of $\mathcal{R}^*$.

- FTC

- $\mathcal R$ is closed under uniform convergence.

**Some bad properties:**

- The Dirichlet function $\indic{x\in\QQ}$ is not in $\mathcal R$.
**(Exercise!)**

  - **Exercise**: Show that $D_f = \RR$ (use sequential continuity)

  - It is in $\mathcal L$ (Lebesgue integral).

- $\mathcal R$ is not closed under pointwise convergence.

  - Example: $g_n(x) = \indic{x\in \QQ,~ x = \frac p q, q \leq n} \in \mathcal R$, but $g_n \not\uniformlyconverges g$. 
  **(Exercise)**

In fact, there exists a sequence of *continuous* functions $\theset{f_n}$ such that

- $0 \leq f_n(x) \leq 1$ for all $x, n$.

- $f_n(x)$ is decreasing as $n\to\infty$ for all $x$.

- $f \coloneqq \lim f_n \not\in\mathcal R$.

This seems disturbing! The Lebesgue integral fixes this particular problem. Letting
$$
\mathcal L = \theset{f \suchthat f \text{ is Lebesgue integrable }},
$$

we have the following theorem:

**Theorem (Dominated Convergence, Special Case):**

Let $\theset{f_n:[a,b] \to \RR} \subset \mathcal L$, such that 
$$
\forall n\in \NN, \forall x, \abs{f_n(x)} \leq M.
$$
If $f_n \to f$ pointwise then $\int f_n \to \int f$.

## Measure Theory


### The Non-Measurable Set
Can we assign a "measure" to all subsets of $\RR^n$?

This should be a function $m: \mathcal P(\RR^n) \to \RR^{\geq 0} \union \theset{\infty} = [0, \infty]$ with some properties (see handout).

- If $\theset{E_i}_{i\in \NN}$ are disjoint, then $m(\disjoint_{i\in \NN} E_i) = \sum_{i\in\NN} m(E_i)$.

- If $E \simeq F$ by translation/rotation/reflection, then $m(E) = m(F)$.

- $m(Q) = 1$ if $q = [0, 1]^n$.

But so far, this is impossible for the following reason:

**Proposition:** 
There exists a non-measurable set:

*Proof:*
Define an equivalence relation $x\sim y \iff x-y \in \QQ$ on $[0, 1)$. Note that each equivalence class bijects with $\QQ$, so each class is countable and there must be an uncountable number of classes. 

Use the axiom of choice to construct a set $N$ by choosing exactly one element from each equivalence class.

Now let $\QQ \intersect [-1, 1] = \theset{q_j}$ be an enumeration of the rationals in this interval, and define $N_j = N + q_j$. 

Note 
$$
j\neq k \implies N_j \intersect N_k = \emptyset
.$$ 
By translation invariance, $m(N_j) = m(N)$, and 
$$
[0, 1) \subseteq \disjoint_j N_j \subseteq [-1, 2]
.$$ 

But then by taking measures and using the fact that $m(N_i) = m(N)$, we have
$$
1 \leq \sum_j m(N_j) \leq 3
,$$ 


But then $m(N) = 0 \implies 1 > m(N)$, and if $m(N) = \varepsilon> 0$ then 
$$
\sum m(N) = \sum \varepsilon > 3.
,$$
a contradiction.
$\qed$

*Exercise:*
Any open set in $\RR$ can be written as a *countable* union of intervals. 

But what can be said about closed sets, or all of $\RR^n$?

**Fact**: 
Any open set in $\RR^n$ can be written as an *almost disjoint* union of closed cubes.

We can then attempt to ascribe a measure to a set by approximating an open set from the inside by cubes.
However, it's not clear that this is unique (although it is).