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[Edited thanks to Daniel's comment.]

The following is the Hahn decomposition theorem stated in Folland's Real Analysis:

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The key step, which could be called as the "greedy algorithm", is as follows

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I would like see how this theorem works for a particular example. Suppose $$ \nu(E):=\int_Ef\ d\mu $$ where $\mu$ is a positive measure on the measurable space $(X,\mathcal{M})$ and $f$ is an extended $\mu$-integrable function.

[Added: one should note that $-\infty$ in the second line of the proof is a typo and it should be $+\infty$ instead.]

Here is my question:

How can one describe the Hahn decomposition for $\nu$ in terms of $f$ and $\mu$? (This is also Exercise 6 in Chapter 3 of Folland's Real Analysis.)

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    Should $\mu \geqslant 0$, or is $\mu$ allowed to be a signed measure? On another note, it seems Folland has a typo in the proof, when he asserts $\nu(P) < \infty$, he must assume that $\nu$ doesn't attain $+\infty$, not that it doesn't attain $-\infty$. – 2017-01-25
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    @DanielFischer: Thanks for your quick comment. I have edited the question accordingly. – 2017-01-25
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    With the clarification that $\mu$ is a positive measure, it seems natural to consider the support of $f^+$ and $f^-$. – 2017-01-25
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    @DanielFischer: In the case that $\mu$ is allowed to be a signed measure, I suppose one needs to consider the decomposition of $\mu$ also? Or is this exercise still doable? – 2017-01-25
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    Pretty much, but $f$ could vanish on a set of positive $\mu$-measure, and we need $P \cup N = X$. One would choose $\{x : f(x) > 0\} \subset P \subset \{ x : f(x) \geqslant 0\}$, usually one of the two extremes. But one can modify by arbitrary null sets. – 2017-01-25
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    Yes, if $\mu$ is signed, one considers the decomposition of $\mu$ too. Or writes $d\mu = h\cdot d\lvert\mu\rvert$ and considers $h\cdot f$ instead of $f$. – 2017-01-25

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If one defines $$ P:=\{x\in X\mid f(x)\geq 0\},\quad N:=\{x\in X\mid f(x)<0\}, $$ one could check that this gives a Hahn decomposition for $\nu$ since $\mu$ is a positive measure.