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Let $\mu$ be a signed measure. I want to prove the following:
(1) If $A$ is a positive set for $\mu$, then $\mu(A)=|\mu(A)|$.
(2) If $A$ is a negative set for $\mu$, then $\mu(A)=-|\mu(A)|$.

This is what I have done:

From $\mu^+ = \frac{1}{2}(|\mu|+\mu)$, I get that $2\mu^+(A)-\mu(A)=|\mu(A)|$.
Similarly, from $\mu^-=\frac{1}{2}(|\mu|-\mu)$, I get $2\mu^-(A)+\mu(A)=|\mu(A)$.

My questions are the following:
Can I say that $\mu^+(A)=\mu(A)$, and $\mu^-(A)=-\mu(A)$, since $A$ is positive in the first case and negative in the second? If so, how can I prove them?

  • 0
    @MichaelGreinecker Maybe you could collect your comments in an answer.2012-02-09

1 Answers 1

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This is really just an issue of looking at the definitions. Let $\mathcal{A}$ be the underlying $\sigma$-algebra.

A positive set $A$ is a measurable set such that $\mu(A)\geq 0$. That is, the function $\mu:\mathcal{A}\to\mathbb{R}$ maps $A$ to a nonnegative number. A number is nonnegative if and only if it equals its own absolute value. So $\mu(A)=|\mu(A)|$ if $A$ is a positive set and that does not require any measure theoretic machinery.

A negative set $A$ is a measurable set such that $\mu(A)\leq 0$. That is, the function $\mu:\mathcal{A}\to\mathbb{R}$ maps $A$ to a nonpositive number. A number is nonpositive if and only if it equals its own absolute value times $-1$. So for a negative set $A$, we have $\mu(A)=-|\mu(A)|$.