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Consider $E\in\mathscr{F}$(sigma-algebra), and $E\in\Omega$ defined on a measure space $(\Omega,\mathscr{F},\mu)$. Suppose $\mu(E)<\infty$, and $\{f_n\}$ is a sequence of measurable functions on $E\to\mathbb{R}$ which are finite almost everywhere and converge almost everywhere to a function $f:E\to\mathbb{R}$ which is also finite almost everywhere. Then $f_n\to f$ almost uniformly in $E$.

Proof: By omitting a subset of $E$ of zero measure, we may assume that all the functions $f_n$ and $f$ are finite that $f_n(x)\to f(x)$ for all $x\in E$.

For positive integers $m,n$ let $$A^m_n=\bigcap_\limits{i=n}^{\infty}\{x:|f_i(x)-f(x)|<\frac{1}{m}\}.$$

Then for fixed $m$, the sequence $A^m_1, A^m_2, \dots,$ is an increasing sequence of measurable sets converging to $E$. Since $\mu(E)$ is finite, by theorem 3.2 there is a positive integer $N_m=N_m(m)$ such that $$\mu(E-A^m_n)<\frac{\epsilon}{2^m}$$ for $n \geqslant N_m$. If we put $$F_\epsilon=\bigcup_\limits{m=1}^{\infty}(E-A^m_{Nm}),$$ then $\mu(F_\epsilon)<\epsilon$. Furthermore, given $\delta>0$, we can choose $m$ so that $\frac{1}{m}<\delta$ and then $|f_i(x)-f(x)|<\delta$ for all $i\geqslant N_m$ $x\in (E-F_\epsilon)$, so that $f_n\to f$ uniformly on $(E-F_\epsilon)$$\blacksquare$ .

I do not understand on this theorem proofs the following points:

  1. Why is $A^m_n=\bigcap_\limits{i=n}^{\infty}\{x:|f_i(x)-f(x)|<\frac{1}{m}\}$ increasing when $m$ is fixed?

  2. How does "$|f_i(x)-f(x)|<\delta$ for all $i\geqslant N_m$ $x\in (E-F_\epsilon)$, so that $f_n\to f$ uniformly" conclude the proof?

  3. What is the conclusion?

Thanks in advance!

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  1. Because the intersection is taking into account less sets, i.e. $$ A^m_n=\bigcap_\limits{i=n}^{\infty}\{x:|f_i(x)-f(x)|<\frac{1}{m}\} \subseteq \bigcap_\limits{i=n+1}^{\infty}\{x:|f_i(x)-f(x)|<\frac{1}{m}\} = A^m_{n + 1}. $$

  2. The result (Egoroff) says, that $f_i$ converges almost uniformly to $f$, i.e. for every $\epsilon > 0$ there exists $F_\epsilon$ with $\mu(F_\epsilon) < \epsilon$ such that convergence on $E - F_\epsilon$ is uniform. But this is just the last sentence of the proof you stated.

  3. should be clear from 2.

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  • $\begingroup$ Why does $f_i$ converges uniformly on $E-F_\epsilon$? What do you define as almost uniform convergence? Thanks for the answer. $\endgroup$ – Pedro Gomes Apr 8 '17 at 11:04
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    $\begingroup$ @PedroGomes For you first question: Because for all $\delta > 0$, there exists $N_m$ such that for all $x \in (E - F_\epsilon)$ and all $i \geq N_m$ you have $|f_i(x) - f(x)| \leq \delta$. This is just the definition of uniform convergence. $\endgroup$ – el_tenedor Apr 8 '17 at 11:18
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    $\begingroup$ @PedroGomes: For your second question see wikipedia. This is the definition I use. But I suppose the author who formulated the theorem in the first place should have stated the definition as well. $\endgroup$ – el_tenedor Apr 8 '17 at 11:18
  • $\begingroup$ Sorry, I explained myself wrong. My first question is: Why is the uniform convergence only on $E-F_\epsilon$ and not on other set? $\endgroup$ – Pedro Gomes Apr 8 '17 at 12:45
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    $\begingroup$ @PedroGomes: The proof does not say anything about other sets, because it is not needed. Just read carefully the definition of "uniform convergence almost everywhere" and compare with the proof and the formulation of Egoroff's theorem. $\endgroup$ – el_tenedor Apr 8 '17 at 13:08

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