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$\hat\mu$ is defined by $$ \hat\mu(E) = \sup \sum_{j=1}^\infty | \mu(A_j)|$$, over all partitions $\{A_j\}$ of $E$.

I have an exercise saying that $\hat\mu \geq \mu$ and that if $\tau \geq \mu$ then $\tau \geq \hat\mu$ (so, $\hat\mu$ is the smallest one that satisfies this).

However, setting $\mu$ as the opposite of the classic Lebesgue measure over the interval $[0,1]$, it's obvious that the measure $\tau \equiv 0$ is smaller than $\hat\mu$ and bigger than $\mu$. Am I missing something?

I can prove the proposition if I change it to $\hat\mu \geq |\mu|$ and that if $\tau \geq |\mu|$ then $\tau \geq \hat\mu$

Does anyone know if that's the real known theorem about total variation measures? Thanks!

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    What is your definition of $|\mu|$? – 2017-02-23
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    Also, in your exampe, does $\geq$ denote absolute continuity? Because in your example, $\mu$ is not absolutely continuous with respect to $\tau$. – 2017-02-23
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    $|\mu|$ is just the absolute value of $\mu$ when applied to a set. And $\geq$ is the good old "bigger than" in $\mathbb{R}$ – 2017-02-23
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    The work is in showing that $\hat \mu $ is a measure; the claim is then immediate. – 2017-02-23
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    @chilango, Yes, of course, but the second claim. I wanted to check that the first one was false – 2017-02-23

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The theorem actually says that if there exists a measure $\nu$ such that $$ \nu(A)\ge|\mu(A)| $$ for any measurable set $A$, then for any such $A$ $$ \nu(A)\ge \hat\mu(A). $$

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    Awesome. Do you, by chance, have a reference? – 2017-02-23
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    You can find the result in many books on real analysis or measure theory books. For instance, Rudin's Real and Complex Analysis. It's on the first 2 pages of chapter 6 on Complex Measures. – 2017-02-23
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    As a side note, your $\hat\mu$ is usually referred to as $|\mu|$ in many textbooks. It is called the totally variation of $\mu$. – 2017-02-23
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    yes, I know, but if I wrote that then the part where I took the absolute value of $\mu$ would have been confusing – 2017-02-23