Question about images of measurable functions

I want to prove the following.

Statement: Given that f is measurable, let
B = {y [tex]\in [/tex] ℝ : μ{f^(-1)(y)} > 0}. I want to prove that B is a countable set.

(to clarify the f^(-1)(y) is the inverse image of y; also μ stands for measure)

Please set me in the right direction. I would greatly appreciate it. Thanks!
 
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Use the fact that any uncountable subset of a complete metric space has a condensation point (a point whose every neighborhood contains uncountably many points of the set).
 
So I've thought about that, but here's the thing. Suppose I assume that B is uncountable. Then, for each y in B, there is a corresponding open image that has measure greater than zero. Since we have uncountably many y's (by assumption), there would be an uncountably many number of open images with measure greater than zero, which implies that the domain is of infinite measure. I'm not sure where to derive the contradiction from this point. Thanks.
 

Dick

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So I've thought about that, but here's the thing. Suppose I assume that B is uncountable. Then, for each y in B, there is a corresponding open image that has measure greater than zero. Since we have uncountably many y's (by assumption), there would be an uncountably many number of open images with measure greater than zero, which implies that the domain is of infinite measure. I'm not sure where to derive the contradiction from this point. Thanks.
Intersect your sets with, say, [0,1] which has finite measure. The sum of the measures of all of the sets in B intersected with [0,1] is less than or equal to 1, right? Can an uncountable number of the measures be positive?
 

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