Let $(X,M,\mu)$ be a finite measure space and let $f$ be a real-valued and measurable function on $X$. How do I evaluate $$ \lim_{n\rightarrow \infty} \int_X ~{\left\{\cos\left(\pi f(x)\right) \right\}}^{2n}~\text{d}\mu(x)$$ in terms of $f$?

Thanks for helping.

  • $\begingroup$ You want the value of the integral or the limit as $n\to\infty$? $\endgroup$ – Julián Aguirre Nov 29 '11 at 10:18
  • $\begingroup$ @JuliánAguirre yes I do. Sorry. I'll edit right away. $\endgroup$ – Colin Nov 29 '11 at 12:27

Put $f_n(x)=\left\{\cos (\pi f(x))\right\}^{2n}$. Then the sequence $\{f_n\}$ is decreasing, and $0\leq f_1\leq 1$, which is integrable. Hence we can apply the reversed version of the monotone convergence theorem, which gives $$\lim_{n\to\infty}\int_X f_n(x)d\mu(x)=\int_X \lim_{n\to\infty}f_n(x)d\mu(x).$$ Since $\left\{\cos(\pi f(x))\right\}^2=1$ if $f(x)\in\mathbb Z$, and $\left\{\cos(\pi f(x))\right\}^2<1$ otherwise, we have $$\lim_{n\to\infty}f_n(x)=\begin{cases}1&\mbox{if }f(x)\in\mathbb Z\\ 0&\mbox{otherwise},\end{cases}$$

so $$\lim_{n\to +\infty}\int_X\left\{\cos (\pi f(x))\right\}^{2n}d\mu(x)=\int_X\mathbf 1_{\left\{f(x)\in\mathbb Z\right\}}d\mu(x)=\sum_{k\in\mathbb Z}\mu(\left\{x\in X,f(x)=k\right\}).$$

  • $\begingroup$ Thanks. Is it supposed to be $0\leq f_n \leq 1$ and $\left\{\cos(\pi f(x))\right\}^{2n}=1$ $\dots$ instead? $\endgroup$ – Colin Nov 30 '11 at 18:16
  • $\begingroup$ Since $f_{n+1}\leq f_n$ for all $n$, we only need that one function of the sequence to be integrable (here in fact all the function are integrable). To see the behaviour of $f_n$, I look first the behaviour of $(\cos (\pi f(x)))^2$. $\endgroup$ – Davide Giraudo Nov 30 '11 at 18:37
  • $\begingroup$ @DavideGiraudo: Please why do we need $\int f_1 \lt \infty$? $\endgroup$ – Kuku Mar 24 '12 at 19:18
  • $\begingroup$ @kuku If not, this version of monotone convergence theorem may be false, for example if you take the real line and $f_n=[n,+\infty)$ then it's a decreasing sequence, each integral is infinite but the integral of the limit is $0$. $\endgroup$ – Davide Giraudo Mar 24 '12 at 19:40
  • $\begingroup$ @DavideGiraudo: Thanks. $\endgroup$ – Kuku Mar 24 '12 at 19:52

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