I am trying to solve the following problem:
Let $f:[0,2)\to \mathbb R$ be defined by
$f(x) = \left\{ \begin{array}{cc} x^2 & \mbox{if } 0\le x \le 1 \\ 3-x & \mbox{if } 1<x <2 \end{array} \right.$
Find $m^*(A)$, where $A=f^{-1}(\dfrac{9}{16},\dfrac{5}{4})$ and $m^*$ denotes the Lebesgue outer measure.
I tried to solve the problem as follows:
Since, $A=f^{-1}(\dfrac{9}{16},\dfrac{5}{4})=(\dfrac{3}{4},1]\cup (\dfrac{7}{4},2)$, thus $m^*(A)\leq \dfrac{1}{2}$. But I could not establish the reverse inequality. Please help!
Well the only thing is to recognize that $A = (\frac 34, 1] \bigcup (\frac 74, 2] \in \mathscr{L}$, where $\mathscr{L}$ denotes the collection of Lebesgue measurable sets. On $\mathscr{L}$, $m^{*}$ actually equals to $m$. So $m^{*}(A) = m(A) = \frac 12$.