Let $(X,M_1,\mu)$ and $(Y,M_2,\phi)$ two measure spaces and a measurable function $f:X \longrightarrow Y$(a function is measurable if $\forall $ $A \in M_2$ we have $f^{-1}(A) \in M_1$)preserves the measure if $\forall $ $A \in M_2$ we have $\mu(f^{-1}(A))=\phi(A)$.
$\{a\}=a-[a]$ the fractional part of $a$.
If we take $X=Y=[0,1)$ and $m$ the lebesgue measure, prove that the function $f(x)=\{2x\}$ preserves the measure.
One idea is to work on the interval $[0,1/2)$ because in this interval $f$ is injective(correct me if i am wrong) and $f(x)=\{2x\}=2\{x\}=2x$ and because of the fact that f is injective we have that $\forall $ $A \subseteq Y$ we have $f^{-1}(f(A))=A$.
Also $2[0,1/2)=[0,1)$
I am stuck of ideas for how to proceed to a complete proof.
Can someone help me or give me a hint?
Thank you!
The proof is straightforward:
First calculate the inverse image of any measurable subset $A\subset [0,1]$. Fairly easy to see that $f^{-1}(A) = A/2 \cup (A/2 + 1/2)$, where the union is disjoint.
Then calculating $$\lambda(f^{-1}(A)) = \lambda(A/2 \cup (A/2 + 1/2)) = \lambda(A)/2 + \lambda(A)/2 = \lambda(A)$$
If you are worried about proving the identity $\lambda(A/2) = \lambda(A)/2$, then you can probably show it works for interval and use measure theoretic induction to prove it works for any subset. Similar argument could be made for the translation invariance use also in the calculation.
Hope this helps!