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On a measure space $(\Omega, \mathcal{F}, \mu)$, in Rudin's Real and Complex Analysis, with some modification of notations:

If there is a set $A \in \mathcal{F}$ such that $\mu(E) = \mu(A \cap E) $ for every $E \in \mathcal{F}$, we say that $\mu$ is concentrated on $A$.

I was wondering if according to the definition, it is true that $\mu$ is concentrated on $A \in \mathcal{F}$ if and only if $\mu(A) = \mu(\Omega) $? If yes, why was the definition made more complicated than what it is?

Thanks and regards!

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up vote 7 down vote accepted

You are right that the condition is too complicated for finite (positive) measures and that in that case your proposed definition is equivalent because $\mu(\Omega) = \mu(A) + \mu(\Omega \smallsetminus A)$, so $\mu(\Omega \smallsetminus A) = 0$ if $\mu$ is finite and positive (hence $\mu$ doesn't charge any subset of $\Omega \smallsetminus A$).

The point is that the measure might be infinite and our argument is no longer valid since we might have used $\infty - \infty =0$, which isn't a good idea. For example, if you think of Lebesgue measure on $\mathbb{R}$ then $\mu([0,\infty)) = \infty$, but it doesn't quite make sense to say that Lebesgue measure is concentrated on $[0,\infty)$, since it charges sets in the negative reals as well. This also illustrates why we don't want to say $\infty - \infty = 0$ in many situations.

Moreover, the definition of Rudin also is the correct one for signed measures.

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Thanks! A further question is about two measures u and v on the same measurable space. u and v are called mutually singular if there exists disjoint measurable subsets A and B, such that u is concentrated on A and v is on B. Are u and v mutually singular if and only if u is zero wherever v is not, and v is zero wherever u is not? – Tim May 12 '11 at 13:53
@Tim: Yes, that's right. – t.b. May 12 '11 at 14:49

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