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I got this exercise as homework and I found some problems in solving it. So I hope that someone can help me.
Let $f:[0,1] \rightarrow R$ Lebesgue measurable and $S=\{x \in [0,1]:f(x) \in Z\}$.
Show that:
$\lambda(S)=\lim_{n \rightarrow \infty} \int_0^1 \lvert \cos(\pi f(x))\rvert^n \,d\lambda$

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4 Answers

up vote 1 down vote accepted

Hint: What is the limit of $\lvert\cos(\pi f(x))\rvert^n$ as $n\to\infty$?

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since $0 \leq \vert \cos{(\pi f(x))}\vert \leq 1$ we have $\vert \cos{\pi f(x)}\vert^n \rightarrow 0$ when $n\rightarrow \infty$ and S has as maximum a numerable cardinality and this means that it has measure zero as union of singletons. –  Laura Nov 12 '12 at 22:25
    
Yes, except when $\cos(\pi f(x))=1$. Which happens when $f(x)=\ldots$? –  Harald Hanche-Olsen Nov 12 '12 at 22:29
    
So $\lambda(S)=0=\lim.\int_0^1 \vert \cos(\pi f(x))\vert^n d\lambda$ always except when it is 1. And it happens when $f(x)=2k,\qquad k\in N$.In this case $\int_0^1 1 d\lambda=1$ –  Laura Nov 12 '12 at 22:32
    
It should be $f(x)=k$. (When $k$ is an odd integer, $\cos(\pi f(x))=-1$, so the absolute value is 1.) But more seriously, your answer cannot possibly depend on $k$, which is just a convenience variable to talk about where the integrand equals $1$. You need to put the integral aside for a moment and focus on the integrand. Do you see why it is $1$ precisely when $x\in S$? And why it goes to zero as $n\to\infty$ for all other values of $x$? –  Harald Hanche-Olsen Nov 13 '12 at 10:08
    
Yes,I mean is $x \in S,\qquad cos(\pi f(x))=+-1,\qquad 1^n \rightarrow 1$ At and if not $-1<cos(\pi f(x))<1$ so $\vert cos(\pi f(x))\vert^n \rightarrow 0$. I think I got this part,I can use it to prove the convergence to che function $X_S$ and then use the monotone convergence as I sum up in my answer that just follows. –  Laura Nov 13 '12 at 13:38
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Try splitting your integral over two sets: $\{\delta<f<1-\delta\}$ and its complement.

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Let $f_n(x):=|\cos(\pi f(x))|$: it's a measurable function, the sequence $\{f_n(x)\}$ is decreasing for all $x$ and converges to $g:=\chi_S(x)$ for all $x$. Apply monotone convergence theorem to $g-f_n$ to get the wanted result.

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I'm going to sum up a complete answer:

$\lambda(S)={x \in [0,1]:f(x) \in Z}$ considering : $\lim_{n \rightarrow \infty} \int_0^1 \vert \cos(\pi f(x))\vert^n d\lambda$ so $0 \leq \vert \cos(\pi f(x))\vert^n \leq 1$. If $\vert \cos(\pi f(x))\vert=1$ it means that $f(x)=2k,2k+1\qquad k \in N \forall x \in [0,1]$ so $\lim_{n \rightarrow \infty} \int_0^1 \vert \cos(\pi f(x))\vert^n d\lambda=\int_0^1 1 d\lambda=1$

At the same time if $f(x)=2k,2k+1\qquad k \in N \forall x \in [0,1]$ it means that $2k,2k+1 \in Z$ and so $\lambda(S)=\lambda([0,1])=1$.

If $\vert\cos(\pi f(x))\vert \neq 1$ you get $\lim_{n \rightarrow \infty} \int_0^1 \vert \cos(\pi f(x))\vert^n d\lambda=\int_0^1 d\lambda=0$ and in this case S has cardinality as maximum numerable so it could be considered as union of singletons of null measure since we get that S has measure zero too.

So I can use this convergence to apply the monotone convergenze theorem to $g-f_n$ as suggested by Davide Giraudo.

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