Showing that $m(\{x \in [0, \pi/2] | \cos(x) \in \mathbb{Q} \}) = m(\{ [0, \pi/2] \cap \mathbb{Q}\}) = 0.$ (if the statement is correct) Showing that $m(\{x \in [0, \pi/2] | \cos(x) \in \mathbb{Q} \}) = m(\{ [0, \pi/2] \cap \mathbb{Q}\}) = 0.$
My Questions are:
for proving $\subseteq$:
Assume that $0 \leq x \leq \pi/2$, then taking the cosine of this, we get $\cos 0 \geq \cos x \geq \cos (\pi/2),$ so, $1 \geq \cos (x) \geq 0$ (because $\cos (x)$ is a decreasing function in this interval.) ..... is my proof correct?
for proving $\supseteq$:
I think that this direction is not true, I think the correct statement should be $m(\{x \in [0, \pi/2] | \cos(x) \in \mathbb{Q} \}) \subseteq m(\{ [0, \pi/2] \cap \mathbb{Q}\}) = 0.$ am I correct? 
Could anyone help me in answering those questions, please?
 A: The two sets are emphatically NOT equal (hint: $x=\pi/3$); the statement only asserts that both sets are measure zero.
The reason the second set is measure zero should be clear using very basic measure theory (presumably you have seen that $\mathbb{Q}$ is measure zero). The reason the first one is measure zero follows by the same reason the second one is. Namely, note that $\cos(x)$ is monotonically decreasing on $[0,\pi/2]$, and so if $f(x)=\cos(x)$, then $f^{-1}(a)$ is either empty or a singleton for each $a\in \mathbb{Q}$. Can you conclude (using the countability of $\mathbb{Q}$ and basic properties of Lebesgue measure)?
A: The other answer covers it, but it may be instructive to note that $\textit{any}$ Lipschitz continuous function maps sets of measure zero to sets of measure zero: 
Suppose for all $x,y\in A,\ |f(x)-f(y)|\le C|x-y|$ for some positive constant $C$. If $A$ has measure zero, we can cover it by a countable union of open balls $\{B_j\}_j$ each of radius $r_j$ such that $\sum |B_j|<\delta$ for any prescribed $\delta>0.$ Then, $f(A)\subseteq \bigcup_j f(B_j)$ and each $f(B_j)$ is contained in a ball of radius less than $Cr_j$ so $|f(A)|\le C\sum |B_j|<C\delta.$
