- #1
futurebird
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I'm learning the proof that [tex]L_{\infty}[/tex] is complete. I do not understand one of the steps.
Let [tex]f_n[/tex] be a cauchy sequence in [tex]L_{\infty}(E)[/tex] then there exists a subset A in E such that [tex]f_n[/tex] is "uniformly cauchy" on E\A. For m,n choose A so that
[tex]|f_n-f_m| \leq ||f_m - f_n||_{\infty}[/tex] for all x in E\A. Take the union of all such As and then [tex]f_n[/tex] converges uniformly on E without the As.
Define f to be [tex]f(x) = \mathop{\lim}\limits_{n \to \infty} f_n(x)[/tex] for x in E without the As, and let it be o otherwise. F is bounded and measurable now all we need is to show that [tex]||f_n - f||_{\infty} \rightarrow 0[/tex] so we know f is in [tex]L_{\infty}[/tex]. We know [tex]m(As)=0[/tex]
This next bit is where the proof makes a leap that I don't understand.
[tex]||f_n - f||_{\infty} \leq \sup_{x \in E-A} |f_n -f|[/tex]
Then is says as n --> infinity we have [tex]\sup_{x \in E-A} |f_n -f| \rightarrow 0[/tex]. But, I have no idea where that inequality came from? What theorem? Please help!
Let [tex]f_n[/tex] be a cauchy sequence in [tex]L_{\infty}(E)[/tex] then there exists a subset A in E such that [tex]f_n[/tex] is "uniformly cauchy" on E\A. For m,n choose A so that
[tex]|f_n-f_m| \leq ||f_m - f_n||_{\infty}[/tex] for all x in E\A. Take the union of all such As and then [tex]f_n[/tex] converges uniformly on E without the As.
Define f to be [tex]f(x) = \mathop{\lim}\limits_{n \to \infty} f_n(x)[/tex] for x in E without the As, and let it be o otherwise. F is bounded and measurable now all we need is to show that [tex]||f_n - f||_{\infty} \rightarrow 0[/tex] so we know f is in [tex]L_{\infty}[/tex]. We know [tex]m(As)=0[/tex]
This next bit is where the proof makes a leap that I don't understand.
[tex]||f_n - f||_{\infty} \leq \sup_{x \in E-A} |f_n -f|[/tex]
Then is says as n --> infinity we have [tex]\sup_{x \in E-A} |f_n -f| \rightarrow 0[/tex]. But, I have no idea where that inequality came from? What theorem? Please help!