(New page: Let <math> f </math> be a non-negative measurable function on <math> \mathbb{R} </math>. Prove that if <math> \sum_{n=-\infty}^{\infty} f(x+n) </math> is integrable, then <math> f=0 </ma...)
 
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[[MA_598R_pweigel_Summer_2009_Lecture_4]]
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Let <math> f </math> be a non-negative measurable function on <math> \mathbb{R} </math>. Prove that if
 
Let <math> f </math> be a non-negative measurable function on <math> \mathbb{R} </math>. Prove that if
  

Revision as of 09:11, 5 July 2009

MA_598R_pweigel_Summer_2009_Lecture_4

Let $ f $ be a non-negative measurable function on $ \mathbb{R} $. Prove that if

$ \sum_{n=-\infty}^{\infty} f(x+n) $

is integrable, then $ f=0 $ a.e.

Proof:

Set $ f_n = \sum_{m=-n}^n f(x+n) $

$ f_n $ are $ L^1 $ since they are measurable and since $ f_n \leq \sum_{n=-\infty}^{\infty} f(x+n) $

Also $ f_n \longrightarrow \sum_{n=-\infty}^{\infty} f(x+n) $

So by Dominated Convergence,

$ \sum_{n=-\infty}^{\infty} \int_{\mathbb{R}} f = lim \int_{\mathbb{R}} f_n =\int_{\mathbb{R}} \sum_{n=-\infty}^{\infty} f(x+n) < \infty $

So $ \int_{\mathbb{R}} f =0 $.

Thus, since $ f \geq 0 $, and $ \int_{\mathbb{R}} f =0 $ , $ f=0 $ a.e.

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