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Thus, since <math> f \geq 0 </math>, and <math> \int_{\mathbb{R}} f =0 </math> , <math> f=0 </math> a.e. | Thus, since <math> f \geq 0 </math>, and <math> \int_{\mathbb{R}} f =0 </math> , <math> f=0 </math> a.e. | ||
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| + | -Jacob Boswell | ||
Latest revision as of 17:12, 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.
-Jacob Boswell
