Math4121 Lecture 21

Without loss of generality, we can assume $f=0$. Given any $\alpha > 0$, $\exists N$ such that

$$\left|\int_0^1 f_n(x) dx \right| < \alpha$$

for all $n\geq N$.

Consider the set $\Gamma_{\alpha/2} = \bigcup_{n=1}^{\infty} E_n$, for each $g\in \Gamma_{\alpha/2}$, we still have $\lim_{n\to\infty}f_n(g) = 0$.

So we define

$$G_i=\{g\in \Gamma_{\alpha/2} :|f_n(g)|<\frac{\alpha}{2} \text{ for all }n\geq i\}$$

So $G_1\subset G_2\subset \cdots$ and $\Gamma_{\alpha/2} = \bigcup_{i=1}^{\infty} G_i$.

By Osgood Lemma, since $\Gamma_{\alpha/2}$ is closed, $\exists K$ such that $c_e(G_K)>c_e(\Gamma_{\alpha/2})-\frac{\alpha}{4B}$.

By definition of $c_e$, we cna find open $I_1,\ldots,I_N$ which cover $\Gamma_{\alpha/2}$ and

$$\sum_{i=1}^N \ell(I_i) < c_e(\Gamma_{\alpha/2})+\frac{\alpha}{4B}$$

Let $\mathcal{U}=\bigcup_{i=1}^N I_i$, and $\mathcal{C}=[0,1]\setminus \mathcal{U}$.

Part 1: Control the integral on $\mathcal{C}$

for each $x\in \mathcal{C}$, $x\notin \Gamma_{\alpha/2}$, so $\exists$ and open interval $I(x)$ and an integer $m(x)$ such that $|f_{m(x)}(x)|<\frac{\alpha}{2}$ and $\forall n\geq m(x), y\in I(x)$

So $\mathcal{C}\subset \bigcup_{x\in \mathcal{C}} I(x)$, and $\mathcal{C}$ is closed and bounded, $\exists x_1,\ldots,x_J$ such that $\mathcal{C}\subset \bigcup_{j=1}^J I(x_j)$. So if $n\geq \max_{j=1,\ldots,J} m(x_j)$, and $x\in \mathcal{C}$, then $|f_n(x)|<\frac{\alpha}{2}$.

So $\int_\mathcal{C} |f_n(x)| dx < \frac{\alpha}{2} c_e(\mathcal{C})$.

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 $\ell(P_2)\leq \frac{\alpha}{4B}$.

Continue on Friday.