3.5 KiB
Math 4121 Lecture 22
Continue on Arzela-Osgood Theorem
Proof:
Part 2: Control the integral on \mathcal{U}
If [x_i,x_{i+1}]\cap G_k\neq \emptyset, then \inf_{x\in [x_i,x_{i+1}]} |f_n(x)| < \frac{\alpha}{2} for all n\geq K. Denote such set as P_1.
Otherwise, we denote such set as P_2.
So \ell(\mathcal{U})=\ell(P_1)+\ell(P_2)\geq c_e(G_K)+\ell(P_2).
This implies \ell(P_2)\leq \frac{\alpha}{4B} since c_e(G_K)\leq c_e(\mathcal{U})+\frac{\alpha}{2B}.
Thus, for n\geq K,
L(P,f_n)\leq \ell(P_1)\frac{\alpha}{2}+\ell(P_2)B
So
\int_\mathcal{U} |f_n(x)| dx \leq c_e(\mathcal{U})\frac{\alpha}{2}+\frac{\alpha}{2}
All in all,
\begin{aligned}
\left\vert \int_\mathcal{U} f_n(x) dx\right\vert &\leq \frac{\alpha}{2}+\frac{\alpha}{2}\\
&= \int_0^1 |f_n(x)| dx\\
&\leq \int_\mathcal{U} |f_n(x)| dx + \int_\mathcal{C} |f_n(x)|dx\\
&\leq c_e(\mathcal{U})\frac{\alpha}{2}+\frac{\alpha}{2}+c_e(\mathcal{C})\frac{\alpha}{2}\\
&= \alpha
\end{aligned}
\forall N\geq K.
QED
Baire Category Theorem
Nowhere dense sets can be large, but they canot cover an open (or closed) interval.
Theorem 4.7 (Baire Category Theorem)
An open interval cannot be covered by a countable union of nowhere dense sets.
Proof:
Suppose (0,1)\subset \bigcup_{n=1}^\infty S_n where each S_n is nowhere dense. In particular, \exists I_1 closed interval such that I_1\subset (0,1) and I_1\cap S_1=\emptyset.
Now for each k\geq 2, S_k is not dense in I_{k-1} so \exists I_k\subsetneq I_{k-1} such that I_k\cap S_k=\emptyset for all j\leq k.
By nested interval property, \exists x\in \bigcap_{n=1}^\infty I_n.
Then x\in (0,1) and x\notin \bigcup_{n=1}^\infty S_n.
Contradiction with the assumption that (0,1)\subset \bigcup_{n=1}^\infty S_n.
QED
Definition First Category
A countable union of nowhere dense sets is called a set of first category.
Corollary 4.8
Complement of a set of first category in \mathbb{R} is dense in \mathbb{R}.
Proof:
We need to show that for every interval I, \exists x\in I\cap S^c. (\exists x\in I and x\notin S)
This is equivalent to the Baire Category Theorem.
QED
Recall a function is pointwise discontinuous if \mathcal{C}=\{c\in [a,b]: f\text{ is continuous at } c\} is dense in [a,b].
\mathcal{D}=[a,b]\setminus \mathcal{C} is called the set of points of discontinuity of f.
Corollary 4.9
f is pointwise discontinuous if and only if \mathcal{D} is of first category.
Proof:
Part 1: If \mathcal{D} is of first category, then f is pointwise discontinuous.
Immediate from Corollary 4.8.
Part 2: If f is pointwise discontinuous, then \mathcal{D} is of first category.
Let P_k=\{x\in [a,b]: w(f;x)\geq \frac{1}{k}\}, \mathcal{D}=\bigcup_{k=1}^\infty P_k.
Need to show that each P_k is nowhere dense. (under the assumption that $\mathcal{C)$ is dense).
Let I\subseteq [a,b] so \exists c\in \mathcal{C}\cap I. So by definition of w(f;c), \exists J\subseteq I and c\in J such that w(f;J)\leq \frac{1}{k} so for all x\in J, w(f;x)\leq \frac{1}{k}. so J\subseteq P_k=\emptyset.
Thus, P_k is nowhere dense.
QED
Corollary 4.10
Let \{f_n\} be a sequence of pointwise discontinuous functions. The set of points at which all f_n are simultaneously continuous is dense (it's also uncountable).
Proof:
\bigcap_{n=1}^\infty \mathcal{C}_n=\left(\bigcup_{n=1}^\infty \mathcal{D}_n\right)^c
The complement of a set of first category is dense.
QED