Every signed measure $\mu$ on $(X,M)$ has a decomposition into difference $\mu=\mu^+ - \mu^-$ , where $\mu^+ , \mu^-$ are measures . (Here $-\infty \lt \mu(E)\le \infty$ for every measurable set $E\in M$ )
I have see Hahn decomposition theorem here . But in Stein & Shakarchi's real analysis Page $_{288}$, it proves this conclusion in another way .
Let $|\mu|$ denote the total variation of $\mu$ , by $$|\mu|(E)=\sup \sum_{j=1}^{\infty}|\mu(E_j)|$$ where the supremum is taken over all partitions of $E$ .
Then we can construct the decomposition as follows : $$\mu^+=\frac12 (|\mu|+\mu)$$ $$\mu^-(E)=0 \,\,\,\,\,\,\,\, \text{whenever $\mu(E)=\infty$}$$ $$\mu^-(E)=\frac12 (|\mu|-\mu)(E) \,\,\,\,\,\,\,\,\,\,\,\, \text{whenever $\mu(E) \lt \infty$}$$ We can see that $\mu^+$ and $\mu^-$ are measures , and they clearly satisfy $$\mu=\mu^+ - \mu^-$$
I can prove that both $|\mu|$ and $\mu^+$ are measures and $\mu \le |\mu|$ . But I have some trouble when dealing with $\mu^-$ .
My attempt:
Remember a set function $v$ called measure iff it satisfy :
$(1)$ For every measurable set $E \in M$ , $0\le v(E) \le \infty$
$(2)$ If $E_1,E_2,... $ is a countable family of disjoint sets in $M$ , then $$\mu(\bigcup_{n=1}^{\infty}E_n)=\sum_{n=1}^{\infty}\mu(E_n)$$
It is easy to see that $\mu^-$ satisfy property $(1)$ . However , if $E_1 , E_2$ denote two disjoint set in $M$ , I can not even prove $\mu^-(E_1 \bigcup E_2) = \mu^-(E_1)+\mu^-(E_2)$ . And I think the example follows might show $\mu^-$ can not be a measure .
Consider the space $R$ with Lebesgue measurable set . Construct a function $f$ : $f(x)=x$ whenever $x\in [-1,\infty)$ and $f(x)=0$ otherwise .
Then we can construct a signed measure $\mu$ by $$\mu(E)=\int_E f(x) \,dx$$
Some simple observation may show that $\mu^-$ defined above can not be a measure . If $\mu^-$ is a measure , then $$\mu^-([-1,2]\bigcup (2,\infty))=\mu^-([-1,2])+\mu^-((2,\infty))$$ Since $\mu([-1,\infty)=\mu((2,\infty))=\infty$ , LHS $=0$ while RHS $=\mu^-([-1,2])$ this may lead to a contradiction if we can prove $\mu^-([-1,2])\gt 0$ . By definition of $|\mu|$ , we have $$|\mu|([-1,2]) \ge |\mu([-1,0]|+|\mu((0,2]|=\frac52$$ . Also , we have $\mu([-1,2]=\frac32$ directly , so $$\mu^-([-1,2])\ge \frac12(\frac52-\frac32)=\frac12$$ which leads to a contradication . So $\mu^-$ defined above can not be a measure .
This example made me so confused , since the decomposition defined above is an theorem in my textbook rather than an exercise . So , did I make something wrong in my discussion above ?