(New page: '''Lemma:''' Let <math>(X,\mathcal{A},\mu)</math> be a finite measure space, and let <math>f</math> be a measurable function on E. Then <math>lim_n \int_X |f|^n = \mu\left\{|f|=1\right\...) |
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Revision as of 08:24, 9 July 2008
Lemma: Let $ (X,\mathcal{A},\mu) $ be a finite measure space, and let $ f $ be a measurable function on E. Then
$ lim_n \int_X |f|^n = \mu\left\{|f|=1\right\} + \infty * \mu\left\{|f|>1\right\} $
where we interpret $ \infty * 0 = 0 $.
Proof: $ \int_X |f|^n = \int_{\left\{|f|>1\right\}} |f|^n + \int_{\left\{|f|=1\right\}} |f|^n + \int_{\left\{|f|<1\right\}} |f|^n $
We have
$ \int_{\left\{|f|=1\right\}} |f|^n = \mu\left\{|f|=1\right\} $,
$ \int_{\left\{|f|>1\right\}} |f|^n = \infty * \mu\left\{|f|>1\right\} $ by Fatou (or Monotone Convergence Theorem), and
$ \int_{\left\{|f|<1\right\}} |f|^n = 0 $ by the Bounded Convergence Theorem, since $ \mu (X) < \infty $. $ \square $
Remark: The hypothesis that $ \mu(X) < \infty $ cannot be omitted, as $ f(x) = \frac{x}{x+1}, X = [0,\infty) $ shows. However, if we require that $ |f|^k \in L^1 $ for some k, then the third equality follows from the Monotone Convergence Theorem, and the Lemma holds under this weaker hypothesis.
a) $ lim_n \int_0^\pi \sin ^n (x)dx = 0 $
b) $ lim_n \int_0^\pi 2^n\sin ^n (x)dx = \infty $
