# Thread: Egoroff's Theorems Proof

1. ## 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 :
$\displaystyle m(A)< \infty$ where $\displaystyle A\subset$\mathbb{R}. Let $\displaystyle f_n$ and $\displaystyle f$ be functions defined on A such that each $\displaystyle f_n \rightarrow f$ almost everywhere on A. Then given any $\displaystyle \epsilon > 0$ there exists a measurable subset $\displaystyle A_\epsilon \subset A$ such that $\displaystyle m(A\backslash A_\epsilon)<\epsilon$ and $\displaystyle f_n \rightarrow f$ uniformly on $\displaystyle A_\epsilon$ .

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

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

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

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

So, MY QUESTIONS ARE :
1) What is this set $\displaystyle E_{m,n} = \bigcap_{i\geq n}$ { $\displaystyle x : |f_i (x)-f(x)|< \frac{1}{m}$ } 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 $\displaystyle n_m$ such that $\displaystyle m(A\backslash E_{m,n_m})$ < $\displaystyle \frac{\varepsilon}{2^m}$ "

3) Where does the properties m(A) < $\displaystyle \infty$ used in the proof?

2. ## Re: Egoroff's Theorems Proof

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

3. ## Re: Egoroff's Theorems Proof

Originally Posted by girdav
1) We write the set on which $\displaystyle f_n$ converges pointwise to f as an union of such sets. We take $\displaystyle \frac 1m$ instead of $\displaystyle \delta\in\mathbb R_+^*$ in order to get a countable family.
2) Since $\displaystyle \lim_{n\to\infty}m(E_{m,n})=0$ for any m, we use the definition of the limit to $\displaystyle \delta=\varepsilon 2^{-m}$ to choose $\displaystyle n_m$ such that $\displaystyle m(E_{m,n_m})\leq \varepsilon 2^{-m}$ (we can choose it to have the inequality $\displaystyle m(E_{m,k})\leq \varepsilon 2^{-m}$ for $\displaystyle k\geq n_m$, but it's not necessary here).
3) We used the fact that m(A)<\infty when we wrote $\displaystyle m(A\setminus E_{m,n})=m(A)-m(E_{m,n})$.
Can the set on which converges pointwise to f be written as $\displaystyle \bigcap_{m\geq1} \bigcup_{n\geq1} \bigcap_{i\geq n}${$\displaystyle x: |f_i(x)-f(x)|\lneq \frac{1}{m}$} ?

4. ## Re: Egoroff's Theorems Proof

Yes, you can see that using the definition of the limit.

5. ## Re: Egoroff's Theorems Proof

Originally Posted by girdav
Yes, you can see that using the definition of the limit.
Then what is difference between $\displaystyle \bigcap_{m\geq1} \bigcup_{n\geq1} \bigcap_{i\geq n}${$\displaystyle x: |f_i(x)-f(x)|\lneq \frac{1}{m}$} AND
$\displaystyle \bigcap_{m\geq1} \bigcap_{i\geq n}${$\displaystyle x: |f_i(x)-f(x)|\lneq \frac{1}{m}$}

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

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# state and prove egoroft theorem

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