Egoroff's Theorems Proof

• Nov 23rd 2011, 12:51 AM
younhock
Egoroff's Theorems Proof
There are some part of this proof I couldn't understand. First, let me state the theorem and its proof which quoted from an analysis book.

EGOROFF's THEOREM :
$m(A)< \infty$ where $A\subset\mathbb{R}$. Let $f_n$ and $f$ be functions defined on A such that each $f_n \rightarrow f$ almost everywhere on A. Then given any $\epsilon > 0$ there exists a measurable subset $A_\epsilon \subset A$ such that $m(A\backslash A_\epsilon)<\epsilon$ and $f_n \rightarrow f$ uniformly on $A_\epsilon$ .

PROOF :
Given that $\varepsilon$>0 and positive integers m and n, define the set
$E_{m,n} = \bigcap_{i\geq n}$ { $x : |f_i (x)-f(x)|< \frac{1}{m}$ } where $i\in \mathbb{N}$ .
If U is the subset on which $f_n \rightarrow f$ , then for any m , $U\subset \bigcup_{n\geq 1}E_{m,n} \subset A$. And Hence we get $m(\bigcup_{n\geq 1} E_{m,n})=m(A)$ .
But, $E_{m,n} \subset E_{m,n+1}$ for all m and n, so $lim_{n\rightarrow \infty} m(A\backslash E_{m,n})$ = $lim_{n\rightarrow\infty}[m(A)-m(E_{m,n})]$= $m(A)-m(\bigcup_{n\geq1}E_{m,n})$ = 0

Thus for each m, there exists an integer $n_m$ such that $m(A\backslash E_{m,n_m})$ < $\frac{\varepsilon}{2^m}$.

Now let $A_\epsilon$ = $\bigcap_{m\geq1}E_{m,n_m}$. Then $A_\epsilon$ is measurable and $m(A\backslash A_\epsilon)$= $m(\bigcup_{m\geq1}(A\backslash E_{m,n_m})) \leq \sum_{m\geq1}m(A-E_{m,n_m})\lneq \sum_{m\geq1}\frac{\epsilon}{2^m}$= $\varepsilon$.

It is easy to show that $f_n \rightarrow f$ uniformly on A.

So, MY QUESTIONS ARE :
1) What is this set $E_{m,n} = \bigcap_{i\geq n}$ { $x : |f_i (x)-f(x)|< \frac{1}{m}$ } defined at the beginning of this proof and why it is defined like this?

2) How can we conclude that " for each m, there exists an integer $n_m$ such that $m(A\backslash E_{m,n_m})$ < $\frac{\varepsilon}{2^m}$ "

3) Where does the properties m(A) < $\infty$ used in the proof?
• Nov 23rd 2011, 01:54 AM
girdav
Re: Egoroff's Theorems Proof
1) We write the set on which $f_n$ converges pointwise to f as an union of such sets. We take $\frac 1m$ instead of $\delta\in\mathbb R_+^*$ in order to get a countable family.
2) Since $\lim_{n\to\infty}m(E_{m,n})=0$ for any m, we use the definition of the limit to $\delta=\varepsilon 2^{-m}$ to choose $n_m$ such that $m(E_{m,n_m})\leq \varepsilon 2^{-m}$ (we can choose it to have the inequality $m(E_{m,k})\leq \varepsilon 2^{-m}$ for $k\geq n_m$, but it's not necessary here).
3) We used the fact that $m(A)<\infty$ when we wrote $m(A\setminus E_{m,n})=m(A)-m(E_{m,n})$.
• Nov 23rd 2011, 08:17 AM
younhock
Re: Egoroff's Theorems Proof
Quote:

Originally Posted by girdav
1) We write the set on which $f_n$ converges pointwise to f as an union of such sets. We take $\frac 1m$ instead of $\delta\in\mathbb R_+^*$ in order to get a countable family.
2) Since $\lim_{n\to\infty}m(E_{m,n})=0$ for any m, we use the definition of the limit to $\delta=\varepsilon 2^{-m}$ to choose $n_m$ such that $m(E_{m,n_m})\leq \varepsilon 2^{-m}$ (we can choose it to have the inequality $m(E_{m,k})\leq \varepsilon 2^{-m}$ for $k\geq n_m$, but it's not necessary here).
3) We used the fact that m(A)<\infty when we wrote $m(A\setminus E_{m,n})=m(A)-m(E_{m,n})$.

Can the set on which converges pointwise to f be written as $\bigcap_{m\geq1} \bigcup_{n\geq1} \bigcap_{i\geq n}${ $x: |f_i(x)-f(x)|\lneq \frac{1}{m}$} ?
• Nov 23rd 2011, 12:06 PM
girdav
Re: Egoroff's Theorems Proof
Yes, you can see that using the definition of the limit.
• Nov 23rd 2011, 11:20 PM
younhock
Re: Egoroff's Theorems Proof
Quote:

Originally Posted by girdav
Yes, you can see that using the definition of the limit.

Then what is difference between $\bigcap_{m\geq1} \bigcup_{n\geq1} \bigcap_{i\geq n}${ $x: |f_i(x)-f(x)|\lneq \frac{1}{m}$} AND
$\bigcap_{m\geq1} \bigcap_{i\geq n}${ $x: |f_i(x)-f(x)|\lneq \frac{1}{m}$}

both also refer to pointwise convergence? or one of them is refer to uniform convergence?