# About the Hahn Decomposition Theorem from J.Yeh Real Analysis.   I must prove that exists a sequence $$\{C_n\}\subseteq \mathcal{A}$$ such that $$\lambda(C_n)\in\mathbb{R}$$ for all $$n\in\mathbb{N}$$, $$\{\lambda(E_n)\}$$ is an increasing sequence and $$\lim_{n\to\infty}\lambda(C_n)=\alpha$$. First, let us suppose that $$\alpha\in [0, +\infty)$$.

Then exists $$C_n\in\mathcal{A}$$ such that $$\alpha-\frac{1}{n}<\lambda(C_n)\le \alpha<\alpha+\frac{1}{n}\quad\forall n\in\mathbb{N},$$ then $$\lambda(C_n)\in\mathbb{R}$$ for all $$n\in\mathbb{N}$$ and $$\lim_{n\to \infty}\lambda(C_n)=\alpha.$$

Let us suppose now that $$\alpha=+\infty$$, then $$\forall m\in\mathbb{N}\quad\exists C_m\in\mathcal{A}\quad\text{such that}\quad \lambda(C_m)> m.\tag1$$ Let be $$M>0$$, then exists $$m\in\mathbb{N}$$ such that $$m>M$$ and therefore from $$(1)$$ exists $$C_m\in\mathcal{A}$$ such that $$\lambda(C_m)>M$$. Thus we have proved that $$\forall M>0,\quad \lambda(C_n)>M\quad \forall n\ge m$$ that is $$\lambda(C_n)\in\mathbb{R}$$ for all $$n\ge m$$ and $$\lim_{n\to\infty}\lambda(C_n)=+\infty$$

In this case however we have only shown that $$\lambda(C_n)\in\mathbb{R}$$ definitively. Is it okay too? Is the sequence $$\{\lambda(C_n)\}$$ increasing?

Could anyone give me some suggestions? Thanks!

• It's very easy to check $\nu(E_n)$ is real in your setup. That was my point; I didn't know why you were worrying about it. And for the specific problem you posted here, it's not really relevant. You're right I made a mistake about $\nu(C_n)\ge\nu(E_n)$, it's not so immediate. I've deleted my answer Sep 13 at 13:14
• Looking at my own notes on the Hahn decomposition, the proof seems to require not only that $(C_n)_n$ is increasing but also that each $C_n$ is a positive set Sep 13 at 13:15
• @FShrikeDo you have the text: Real analysis Theory of Measure and Integration (3rd Edition) of J.Yeh? If yes, you look at page 219 Theorem 10.14. In any case, I need that $\nu(C_n)\in\mathbb{R}$ Sep 13 at 13:19
• I didn't study from that text, no. Sep 13 at 13:20

I don't know what you mean when you talk about $$\nu(E_n)\in\Bbb R$$ since $$\nu$$ is by definition not going to attain $$+\infty$$ and you know $$\nu(E_n)$$ is nonnegative.

The concept of positive set is very key. Suppose $$\alpha<\infty$$ for the moment. If $$\alpha=0$$ this is very trivial (as is the statement of the Hahn decomposition theorem), so take $$\alpha>0$$. Start by choosing any $$E_1$$ with $$\nu(E_1)>0$$ and $$\alpha-\nu(E_1)<1$$. It's not a super obvious lemma but it is true that $$E_1$$ must contain a positive subset $$C_1$$; meaning $$\nu(K)\ge0$$ for all measurable $$K\subseteq C_1$$ and $$\nu(C_1)>0$$. More can be said; we can choose $$C_1$$ such that $$\nu(C_1)\ge\nu(E_1)$$. In particular, $$0\le\alpha-\nu(C_1)<1$$.

Similarly we can find a positive $$C_2$$ such that $$0\le\alpha-\nu(C_2)<1/2$$. And for general $$n\in\Bbb N$$ we can find a positive $$C_n$$ such that $$0\le\alpha-\nu(C_n)<1/n$$; now $$C'_n:=\bigcup_{j=1}^nC_j$$ will suffice.

Because each $$C'_n$$ is a positive set (by finite induction) and it follows (why?) that $$\nu$$ is an ordinary measure when restricted to positive subspaces - in particular we have monotonicity - $$\nu(C_n)\le\nu(C'_n)$$ and $$\lim_n\nu(C_n)=\alpha$$ thus $$\lim_n\nu(C'_n)=\alpha$$ too. Of course $$(C'_n)_n$$ is an increasing sequence.

So we have an increasing sequence of positive sets whose measures converge to $$\alpha$$. In the $$\alpha=\infty$$ case you do a similar thing.

• Thanks for answer, but this doesn't solve the problem. For some previous steps in the theorem it is necessary that $\nu(E_n)\in\mathbb{R}$ for all $n\in\mathbb{N}$. Why $\nu(C_n)\ge \nu(E_n)$? $\nu$ is a signed measure Sep 13 at 13:04
• Fixed, I think @NatMath Sep 13 at 13:49
• @FShrikeFor greater clarity I have posted the part of the theorem that interests us. Sep 13 at 16:02