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 $\nu(E)<+\infty$, 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?
 A: Let $(X, \mathcal{M})$ be a measurable space and $\nu$ be a signed measure on $\sigma$-algebra $\mathcal{M}$ such that $\forall E \in \mathcal{M}, \nu (E) \in \mathbb{R}$. Suppose that $\nu$ is not bounded.
Then we have a sequence $(E_{n})_{n \in \mathbb{N}} \in \mathcal{M}^{\mathbb{N}}$ such that $\lim_{n \to +\infty} |\nu (E_{n})| = +\infty$. Since $\nu (E_{n}) + \nu (X \setminus E_{n}) = \nu (X)$ and since $\nu(E_{n}), \nu(X\setminus E_{n}), \nu(X) \in \mathbb{R}$, so we can suppose that $\lim_{n \to +\infty} \nu (E_{n}) = +\infty$ by eventually replacing $E_{n}$ by $X \setminus E_{n}$. By taking a subsequence of $(E_{n})_{n \in \mathbb{N}}$, we suppose from now on that $\nu(E_{n}) \geq 2^{n-1}n$ for every $n \in \mathbb{N}$.
Denote $E^{-1} := X \setminus E, E^{1} := E$ for every $E \subset X$, and for every $n \in \mathbb{N}$, $A_{n} := \{\cap_{k = 1}^{n} E_{k}^{c_{k}} \mid (c_{1}, \ldots, c_{n}) \in \{-1, 1\}^{n} \setminus \{(-1, \ldots, -1)\}\}$, $D_{n} \in A_{n}$ such that $\nu(D_{n}) = \max \{\nu(S) \mid S \in A_{n}\}$. Then for every $n \leq m$  in $\mathbb{N}$, $2^{n-1} \nu(D_{n}) \geq \mu (E_{n}) \geq 2^{n-1}n$ and $D_{n} \supset D_{m}$ or $D_{n} \cap D_{m} = \emptyset$.
Denote $A = \{D_{n} \mid n\in \mathbb{N} ~ \forall k > n, D_{n} \cap D_{k} = \emptyset\}$.
If $\mathrm{Card} A < \infty$, we have a subsequence $(D_{\sigma(k)})_{k \in \mathbb{N}}$ of $(D_{n})_{n \in \mathbb{N}}$ which is decreasing for inclusion, so $+\infty = \lim_{k \to +\infty} \nu (D_{\sigma(k)}) = \nu (\cap_{k \in \mathbb{N}} D_{\sigma(k)})$, absure.
If $\mathrm{Card} A = \infty$, we have a subsequence $(D_{\sigma(k)})_{k \in \mathbb{N}}$ of $(D_{n})_{n \in \mathbb{N}}$ where $D_{\sigma(k)}, k \in \mathbb{N}$ are pairwise disjoints, so $+\infty = \sum_{k \in \mathbb{N}} \nu (D_{\sigma(k)}) = \nu (\sqcup_{k \in \mathbb{N}} D_{\sigma(k)})$, absure.
A: 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 $\nu(E_n)>0$ and by taking differences of sets of a subsequence for which $\nu(E_{n_k})>2^k$ 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 $\nu(E_n)\to-\infty$.
