I am trying to solve Exercise 3.21 in Folland's real analysis. I am watching the solution to this question in the following video.
https://www.youtube.com/watch?v=E8FwgGn1e_o
At about 2:31 in the video he says something along the lines of, "we can bring the sum inside the modulus since the sets are disjoint"
That is to say that if $E_1, E_2,...$ are disjoint measurable sets such that $E=\cup_{i=1}^{\infty}E_i$ then
$\sum_{i=1}^{\infty}\lvert \nu (E_i) \rvert =\sum_{i=1}^{\infty}\lvert \int_{E}\chi_{E_i}d\nu \rvert=\lvert \int_{E}(\sum_{i=1}^{\infty}\chi_{E_1})d\nu\rvert$. The first equality is fine but the second equality does not seem true to me. At the end of the day those integrals are complex numbers, so it does not seem reasonable that this should be true. Note that $\nu$ is a complex measure.
The reasoning is definitely not correct, or at least not worded properly. But we can amend this as follows:
Given $E_1,E_2,\ldots$ disjoint with $E=\cup_jE_j$, define $g:X\to\mathbb C$ by $$g=\sum_{j=1}^\infty\overline{\operatorname{sgn}(\nu(E_j))}\chi_{E_j}.$$ Then $|g|\leq1$, and $$\sum_{j=1}^\infty|\nu(E_j)|=\left|\int_Eg\ d\nu\right|\leq\sup\left\{\left|\int f\ d\nu\right|:|f|\leq 1\right\}.$$ Taking the supremum over all such partitions $\{E_j\}$ of $E$, we obtain $\mu_2(E)\leq\mu_3(E)$.