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 $(f^{-1}(A)
\setminus B)\subset [a,b].$

I can see why intuitively this seems to be false : if you just take a function that goes above $1$ with some pulses and that the length of those pulses decrease geometrically but are always of length strictly positive, the measure of the pre-image will be finite but you will always have a non-zero measure for $f^{-1}(A) \backslash [a,b]$, hence $B$ cannot be found with measure zero.
Constructing an explicit example of that kind of thing isn't usually very satisfying, but it needs to be done for confirmation. I'll try thinking of one and edit this answer. 
Hope that helps,
EDIT : I guess Byron's function is working just fine.
