3
$\begingroup$

Suppose that $A \subset [0, 1]$ is a measurable set. Prove that the set $B = \cos A := \{\cos x: x \in A\}$ is measurable and $\mu(B) \leq 0.85\mu(A)$, where $\mu$ is the Lebesgue measure.

  • 4
    Please tell us what you have managed to do so far, and the context in which this arises. This will allow people to give you answers that don't repeat what you already know (thereby wasting their time and your time) and also to know at what level to write the answer. Simply quoting your homework problem and saying "Thank you very much" does not help anybody figure out why you are having trouble with this.2012-02-20

1 Answers 1

3

To see that $B$ is measurable, note that $\cos x$ has a continuous inverse $\arccos(x)$ on the interval $[0,1]$, hence $B=\arccos^{-1}(A)$ which is the pre-image of a measurable set under a continuous function so $B$ is measurable.

On the other hand, try to see what happen to the intervals. Let $A=[a,b]\subset [0,1]$ then $m(A)=b-a$ and $B=[\cos(b),\cos(a)]$ so $m(B)=|\cos(b)-\cos(a)|=|\sin(\xi)|b-a|$ for some $\xi\in (a,b)$ (the last equality follows from the mean value theorem). Now, $\sin(x)$ is increasing on the interval $[0,1]$ so $|\sin(\xi)|\leq \sin(1)\leq .85$. Thus, $m(B)\leq .85 m(A)$ (in this case). You can now use the definition of Lebesgue measure to extend the result to an arbitrary measurable set.

  • 0
    I don't know how to use the definition of Lebesgue measure to extend the result to an arbitrary measurable set.2012-02-21