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

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up vote 4 down vote accepted

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

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Plese see the revised version – jenny Nov 10 '11 at 20:51
thank you. you are correct. – jenny Nov 10 '11 at 21:09
Heh, Byron, was that synchronism. And the same ideas on top of that. – Patrick Da Silva Nov 10 '11 at 21:15
+1 for using the distance-to-F function to define concisely $f$. – Did Nov 10 '11 at 22:27

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.

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