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