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Suppose we have$ (R, P(R), v)$, is measurable space where, $R$-real number, $P(R)$ is power set, $v$, counting measure.

Consider $ f(x)=1/2^x$, let, $E=[0,1]$.

$$\int_E f(x)\text{d}(v)$$ is infinite. In general . What can we say for arbitrary function when integrate it with respect to counting measure on uncountable set.

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In general: if $\int_E f(x)\,d \nu$ is finite and $\nu$ denotes the counting measure, then $\{x: f(x) \neq 0\}$ is at most countable.

Hint: Note that $f$ is inegrable iff $f^+$ and $f^-$ are integrable. Note also that $$ \{x: f(x) > 0\} = \bigcup_{n=1}^\infty \{x: f(x) \geq 1/n\} $$

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  • $\begingroup$ Thnaks, could give a hint how can I prove it? $\endgroup$
    – Team
    Feb 8, 2017 at 15:34
  • $\begingroup$ One more question, in general, if we integrate for the same measurable space but we change our measure as following : $$ v(A)=0$$ for countable set and $$v=infinite $$ otherwise. Is there any Lemma like the one that mentioned $\endgroup$
    – Team
    Feb 8, 2017 at 15:46
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    $\begingroup$ @Team for your measure, every integral is either zero, infinite, or non-existent (as in $\infty - \infty$). $\endgroup$
    – Ben Grossmann
    Feb 8, 2017 at 15:49
  • $\begingroup$ you are help me a lot . just a hint why. since it is first time I deal with this topic $\endgroup$
    – Team
    Feb 8, 2017 at 15:59
  • $\begingroup$ @Team show that this is true for the integral of any "simple function". It follows that the same holds for arbitrary functions. $\endgroup$
    – Ben Grossmann
    Feb 8, 2017 at 16:01

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