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Let $dx$ be the Lebesgue measure on $\mathbb R^d$. Let $u:\mathbb R^d\to{\mathbb R}\cup\{\infty\}$ is a non-negative and measurable function. The question is that, what are the conditions on $u$ so that $u$ is a density of a positive measure $\mu$, that is $d\mu(x)=u(x)dx$.If $u$ is finite on a positive measurable $E$, then is it true that $u$ is a density of a positive measure ?

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I think that any $u$ will do. You always have finite additivity for $$\mu(A)=\int_A u(x)\, dx$$ and you can easily check denumerable additivity by monotone convergence (of course, here positivity of $u$ is essential). –  Giuseppe Negro Jun 9 '11 at 10:02
    
Do you mean here: the measure of the set $\{x:u(x)>0\}$ is positive? –  Hai Minh Jun 9 '11 at 10:13
    
dissonance is right, writing $d \mu (x) := u(x) d x$ defines a positive measure. –  Tim van Beek Jun 9 '11 at 10:31
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Indeed. Even if $u(x) = \infty$ for all $x$ you get a measure, it's just not very interesting. –  Mark Jun 9 '11 at 10:47
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Provided $u$ is measurable. –  Did Jun 9 '11 at 11:02

1 Answer 1

up vote 2 down vote accepted

As explained in the comments, every (measurable) function $u$ will do.

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