About the Hahn Decomposition Theorem from J.Yeh Real Analysis.

Question

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!

Answer 1

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.