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Folland in his book "Real Analysis" says

The real and imaginary parts $\nu_r, \nu_i$ of complex measures $\nu$ are signed measures which do not assume the values $+\infty$ or $-\infty$; hence they are bounded

But, I don't understand why the finiteness of $\nu_r,\nu_i$ means their boundedness. Why is this the case?

For example, the function $1/x$ on $(0,3)$ is finite valued, though it is unbounded.

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    $\begingroup$ See Theorem 6.4 in Rudin's Real and Complex Analysis. $\endgroup$ – fourierwho Apr 13 '18 at 17:41
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It is hidden in the definition of $\sigma$-additivity. Since the property holds for any sequence of measurable sets, you can take the same collection and permute them in any way. Since the property claims that the sum converges, regardless of the permutation, therefore it is claiming absolute convergence (by Riemann's theorem).

Therefore, if you had a sequence of sets such that their measures' real or imaginary part tend to infinity, you can first replace them by a sequence of disjoint measurable sets with the same property. Then that new sequence would contradict the $\sigma$-additivity applied to them and their union. the reason is that their union has as measure the sum of the series. It is part of the definition of complex measure that the values of the measure are complex numbers. In particular, no $\infty$, as for positive measure. Therefore, the sum above cannot have a real or imaginary part that is infinite.

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