A problem about measure Let $f\in C[0,\infty)$. Define $A=[1,\infty)$ and $f^{-1}(A)=\{s\in [0,\infty): f(s)\geq 1\}$. If the measure $m(f^{-1}(A))<\infty$, then there exists an bounded interval $[a,b]\subset [0,\infty)$ and a measurable $B\subset[0,\infty)$ with $m(B)=0$ such that $(f^{-1}(A)
\setminus B)\subset [a,b].$

Updated answer to new problem: This is still false.
Let $F$ be the closed set $\cup_{n=1}^\infty [n,n+1/2^n]$ and set 
$f(x)=1-d(x,F)$, where $d(x,F)$ is the distance from $x$ to $F$.  Then $f(x)=1$ 
if and only if $x\in F$, so $f^{-1}(A)=F$. 

Try $f(x)=\sin(x)$. ${}{}{}{}{}{}{}{}$
