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quasar987

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## Homework Statement

I would like to show that if we have a non-negative real valued function f defined on f a set X, and that the set of points where f is non-vanishing is uncountable, then for any M > 0, I can find a sequence {x_n} of points in X such that

[tex]\sum_n f(x_n)>M[/tex]

## Homework Equations

## The Attempt at a Solution

This is not for a set theory class, its in the context of a measure and integration problem where I want to demonstrate a necessary and sufficient condition for a function to be integrable on a set X with respect to the cardinal measure mu(E) = |E|.

Ok, so I would appreciate to get a confirmation that the following assertion is true, because if it is, then I will have solved the problem I think:

"If for any natural number n,

[tex]f^{-1}(]\frac{1}{n}, +\infty[)[/tex]

is countable, then

[tex]f^{-1}(]0, +\infty[)[/tex]

is countable too."

It seems so to me since it seems natural that a countable union of countable sets in countable and

[tex]\bigcup_{n\geq 1} f^{-1}\left(]\frac{1}{n}, +\infty[\right) = f^{-1}\left(\bigcup_{n\geq 1}]\frac{1}{n}, +\infty[\right)=f^{-1}(]0, +\infty[)[/tex]

Thx.