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 am revisiting the Ergoroff Theorem and I have a deep doubt:

How can we go from a function that converges almost everywhere from a function that converges uniformly on $E-F_\epsilon$? I mean the function is only assumed to be pointwise convergent. How can a pointwise convergent function on a set $E$ become uniformly convergent on a set $E-F_\epsilon$(given the fact $\mu(F_{\epsilon}<\epsilon\:\:for\:\epsilon>0)$?

Thanks in advance!

  • $\begingroup$ The uniform convergence on $E \setminus F_\epsilon$ comes, of course, from the definition of lying in all the $A_{N_m}^m$ $\endgroup$ – Bananach May 29 '17 at 15:32
  • $\begingroup$ Exercise: Show that $E\setminus F_{\epsilon}=\bigcap_{m=1}^{\infty}A_{N_m}^{m}$ $\endgroup$ – Bananach May 29 '17 at 15:33
  • $\begingroup$ @Bananach Sorry but I did not understand your first comment. $\endgroup$ – Pedro Gomes May 29 '17 at 15:40
  • $\begingroup$ There wasn't much to understand from my first comment. Why don't you try and prove uniform convergence on $E\setminus F_{\epsilon}$. Or at least write out what you need to show for this exactly, and which point you have trouble proving that? $\endgroup$ – Bananach May 29 '17 at 17:08
  • $\begingroup$ My problem is that I cannot exclude the uniform convergence in the $E -F_\epsilon$ as an assumption made by the proof. Once we set off a function that is only assumed to converge point wise. I wonder how is it possible for that same function to converge uniformly on so large set $E-F_\epsilon$ compared to a small set $F_\epsilon$? $\endgroup$ – Pedro Gomes May 29 '17 at 17:13

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