Math4121 Lecture 15

Continue on patches for Riemann integral

Hankel's conjecture

If f is pointwise discontinuous (set of points of continuity is dense), then f\in\mathcal{R}.

Why the conjecture is believable:

\{w(f)\leq \sigma\}=\bigcup_{a\in D} I_a, so then S_{\sigma} contains no intervals, so the outer content maybe zero.

However, it turns out that the complement of the set of points of continuity can be large.

Definition: Accumulation point

Given a set S, x is an accumulation point of S if every open interval containing x also contains infinitely many points of S.

The derived set of S, denoted S', is the set of all accumulation points of S.

Lemma


c_e(S)\leq c_e(S')

Proof:

Given \epsilon > 0, we can find open intervals I_1, I_2, \ldots such that S'\subseteq \bigcup_{i=1}^{n} I_i and


\sum_{i=1}^{n} \ell(I_i) \leq c_e(S') + \frac{\epsilon}{2}

So S\setminus \bigcup_{i=1}^{n} I_i contains only finitely many points, say N, so we can cover S\setminus \bigcup_{i=1}^{n} I_i by N intervals of length \frac{\epsilon}{2N}. We call these intervals J_1, J_2, \ldots, J_N. Then S is covered by the intervals C=\bigcup_{i=1}^{n} I_i \cup \bigcup_{i=1}^{N} J_i. and \ell(C)\leq c_e(S') + \frac{\epsilon}{2} + \frac{N\epsilon}{2N}=c_e(S')+\epsilon.

So c_e(S)\leq \ell(C)\leq c_e(S')+\epsilon.

QED

Corollary: sef of first species

There exists n\in \mathbb{N} such that S^{(n)'}=\phi.

If S is of the first species, then c_e(S)=0. This leads to further credence to Hankel's conjecture, but it begs the question is the complement of \bigcup_{a\in D} I_a indeed of first species?

Theorem (Baire Category Theorem)

Every open set in \mathbb{R} is a countable union of disjoint open intervals.

Proof:

Let U be open and for each t\in U, set a=\inf \{x:(x,t]\subseteq U\} and b=\sup \{x:[t,x)\subseteq U\}.

We define I(t)=(a,b), U\subseteq \bigcup_{t\in U} I(t).

Notice that if I(t)\cap I(s)\neq \emptyset, then there exists rational r\in U such that I(t)\cap I(s)\subseteq (r,r).

Take a dense, contable set in [0,1], say D=\{a_n\}_{n=1}^{\infty}. Let \epsilon>0 and define I_n=(a_n-\frac{\epsilon}{2^{n+1}}, a_n+\frac{\epsilon}{2^{n+1}}). Can the entire interval [0,1] be placed inside a union of intervals whose lengths add up to \epsilon?


\sum_{n=1}^{\infty} \ell(I_n) = \sum_{n=1}^{\infty} \frac{2\epsilon}{2^{n+1}} = \epsilon

Harnack believed that the complement of a countable union of intervals is another countable union of intervals.

Borel found the flaw in Harnack's argument. In fact, [0,1]\setminus \bigcup_{n=1}^{\infty} I_n is uncountable.

Theorem (Heine-Borel)

If \{U_\alpha\}_{\alpha\in A} is a collection of open sets (countable or uncountable) such that


[a,b]\subseteq \bigcup_{\alpha\in A} U_\alpha

then there exists a finite or countable subcollection \{U_{\alpha_n}\}_{n=1}^{\infty} such that


[a,b]\subseteq \bigcup_{n=1}^{\infty} U_{\alpha_n}

Continue on next lecture.

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