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'''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 | '''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> | + | <math>\lim_{n \rightarrow \infty} \int_X |f|^n = \mu\left\{|f|=1\right\} + \infty * \mu\left\{|f|>1\right\}</math> |
where we interpret <math>\infty * 0 = 0</math>. | where we interpret <math>\infty * 0 = 0</math>. | ||
| − | '''Proof:''' <math>\int_X |f|^n = \ | + | '''Proof:''' <math>\int_X |f|^n = \int\limits_{\left\{|f|>1\right\}}\!\!\!\!\!|f|^n + \!\! \int\limits_{\left\{|f|=1\right\}}\!\!\!\!\!|f|^n + \!\! \int\limits_{\left\{|f|<1\right\}}\!\!\!\!\!|f|^n </math> |
We have | We have | ||
| − | <math>\ | + | <math>\int\limits_{\left\{|f|=1\right\}} \!\!\!\!\! |f|^n \rightarrow \mu\left\{|f|=1\right\} </math>, |
| − | <math>\ | + | <math>\int\limits_{\left\{|f|>1\right\}} \!\!\!\!\!|f|^n \rightarrow \infty * \mu\left\{|f|>1\right\} </math> by Fatou (or Monotone Convergence Theorem), and |
| − | <math>\ | + | <math>\int\limits_{\left\{|f|<1\right\}} \!\!\!\!\! |f|^n \rightarrow 0 </math> by the Bounded Convergence Theorem, since <math>\mu (X) < \infty</math>. <math>\square</math> |
'''Remark:''' The hypothesis that <math> \mu(X) < \infty </math> cannot be omitted, as <math>f(x) = \frac{x}{x+1}, X = [0,\infty)</math> shows. However, if we require that <math>|f|^k \in L^1</math> for some k, then the third equality follows from the Monotone Convergence Theorem, and the Lemma holds under this weaker hypothesis. | '''Remark:''' The hypothesis that <math> \mu(X) < \infty </math> cannot be omitted, as <math>f(x) = \frac{x}{x+1}, X = [0,\infty)</math> shows. However, if we require that <math>|f|^k \in L^1</math> for some k, then the third equality follows from the Monotone Convergence Theorem, and the Lemma holds under this weaker hypothesis. | ||
| − | a) <math> | + | a) <math>\lim_{n \rightarrow \infty} \int_0^\pi \sin^n(x)dx = 0</math> |
| − | b) <math> | + | b) <math>\lim_{n \rightarrow \infty} \int_0^\pi 2^n\sin^n(x)dx = \infty</math> |
Latest revision as of 09:55, 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 \rightarrow \infty} \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\limits_{\left\{|f|>1\right\}}\!\!\!\!\!|f|^n + \!\! \int\limits_{\left\{|f|=1\right\}}\!\!\!\!\!|f|^n + \!\! \int\limits_{\left\{|f|<1\right\}}\!\!\!\!\!|f|^n $
We have
$ \int\limits_{\left\{|f|=1\right\}} \!\!\!\!\! |f|^n \rightarrow \mu\left\{|f|=1\right\} $,
$ \int\limits_{\left\{|f|>1\right\}} \!\!\!\!\!|f|^n \rightarrow \infty * \mu\left\{|f|>1\right\} $ by Fatou (or Monotone Convergence Theorem), and
$ \int\limits_{\left\{|f|<1\right\}} \!\!\!\!\! |f|^n \rightarrow 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 \rightarrow \infty} \int_0^\pi \sin^n(x)dx = 0 $
b) $ \lim_{n \rightarrow \infty} \int_0^\pi 2^n\sin^n(x)dx = \infty $
