2.5 KiB
Math4121 Lecture 21
Rolling from last lecture
Convergence of integrals
Arzela-Osgood Theorem
Let \{f_n\} be a sequence of function, f(x)=\lim_{n\to\infty}f_n(x) for every x\in [0,1], if f\in \mathscr{R}[0,1], and \exists B>0 such that |f_n(x)|\leq B \forall x\in [0,1]. (uniformly bounded and integrable)
\lim_{n\to\infty}\int_0^1 f_n(x) dx = \int_0^1 f(x) dx
If we let \Gamma_{\alpha} be the set of intervals where f_n is not continuous,
\Gamma_{\alpha} = \{x\in [0,1] : \textup{ for any }m\in \mathbb{N}, \delta > 0, \exists n\geq m, y\in (x-\delta, x+\delta) \text{ s.t. } |f_n(y)-f(y)|>\alpha\}
Fact: \Gamma_{\alpha} is closed and nowhere dense.
Proof
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.