A signed measure is bounded If $\nu$ is a signed measure defined over $(X,\mathcal{M})$ such that $\nu(E)\in\Bbb{R}$, for all $E\in\mathcal{M}$, then $\nu$ is bounded.


This looks weird to me. Of course, $\nu(E)\in\Bbb{R}$ implies that $\nu(E)&lt;+\infty$, once we defines a signed measure to take values at $\Bbb{R}\cup\{-\infty,+\infty\}$. But only with this informations, how conclude that $\nu$ is bounded?
Assume that $E_1,E_2,...$ satisfy that $\nu(E_n)\to+\infty$. By excluding finitely many terms from the beginning, we can assume that $\nu(E_n)&gt;0$ and by taking differences of sets of a subsequence for which $\nu(E_{n_k})&gt;2^k$ that the $E_n$ are dijoint. Then $\mathbb{R}\ni\nu\left(\bigcup_nE_n\right)=\sum_{n=1}^{\infty}\nu(E_n)=+\infty$. Therefore, there is no such sequence. 

Similarly, you can exclude the existence of a sequence with $\nu(E_n)\to-\infty$.
Assume that $E_1,E_2,...$ satisfy that $\nu(E_n)\to+\infty$. By excluding finitely many terms from the beginning, we can assume that $\nu(E_n)&gt;0$ and by taking differences of sets of a subsequence for which $\nu(E_{n_k})&gt;2^k$ that the $E_n$ are dijoint. Then $\mathbb{R}\ni\nu\left(\bigcup_nE_n\right)=\sum_{n=1}^{\infty}\nu(E_n)=+\infty$. Therefore, there is no such sequence. 

Similarly, you can exclude the existence of a sequence with $\nu(E_n)\to-\infty$.