Math 4121 Lecture 22

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,

$$\begin{aligned} \left\vert \int_\mathcal{U} f_n(x) dx\right\vert &\leq \frac{\alpha}{2}+\frac{\alpha}{2}\\ &= \int_0^1 |f_n(x)| dx\\ &\leq \int_\mathcal{U} |f_n(x)| dx + \int_\mathcal{C} |f_n(x)|dx\\ &\leq c_e(\mathcal{U})\frac{\alpha}{2}+\frac{\alpha}{2}+c_e(\mathcal{C})\frac{\alpha}{2}\\ &= \alpha \end{aligned}$$

$\forall N\geq K$.

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 $(0,1)\subset \bigcup_{n=1}^\infty S_n$.

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.

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.

$$\bigcap_{n=1}^\infty \mathcal{C}_n=\left(\bigcup_{n=1}^\infty \mathcal{D}_n\right)^c$$

The complement of a set of first category is dense.