# Existence of sequence converging to the supremum over all measuable sets in $(\Omega, \mathcal{F})$.

$$\newcommand{\scrF}{\mathcal{F}}$$ Theorem: Let $$\mu$$ be a signed measure on $$(\Omega, \mathcal{F})$$. Then there exists $$E^* \in \mathcal{F}$$ and $$E_* \in \mathcal{F}$$ such that $$\mu(E^*) = \sup_{E \in \mathcal{F}} \mu(E) \text{ and } \mu(E_*) = \inf_{E \in \mathcal{F}} \mu(E).$$ My issue is with the first step of the proof unfortunately, and I think it's fairly basic. We let $$L = \sup_{E \in \mathcal{F}}\mu(E)$$, then apparently there exists some sequence $$(E_n)$$ in $$\mathcal{F}$$ such that $$\lim_{n\rightarrow \infty} \mu(E_n) = L$$. But why do we know such a sequence exists? Sorry I'm sure it's basic I just don't see it, thanks in advance.

Edit: Aparently this follows from basic properties of liminf and limsup, but I still don't see it? If you think there should be more context/ information please let me know! I know there is no assumption that $$\mu$$ is $$\sigma$$-finite.

Edit 2: Maybe some definitions:

Definition: We say a function $$\mu:\scrF \rightarrow (-\infty,\infty]$$ is a signed measure on $$\scrF$$ satisfying $$\mu(\varnothing) = 0$$ and for any sequence of disjoint sets $$(E_n) \in \scrF$$ we have that $$\mu \left( \bigcup_{n=1}^\infty E_n\right)= \sum_{n=1}^\infty \mu(E_n).$$

So I'm wondering if this is just a supremum and infimum situation? Because I guess we can just say that $$S = \{\mu(A) : A \in \scrF \}$$, this is a set of real numbers and we know by the axiom of completeness that $$\sup S$$ exists so $$\sup_{E \in \scrF} \mu(E) = \sup S$$ exists? I know that if we have some arbitrary set $$B \subseteq \mathbb{R}$$ that there is a sequence in $$\mathbb{R}$$ converging to the supremum, but in this case directly applying this would give me a sequence not necessarily in the sigma-field $$\scrF$$ right? Am I just overthinking this?

• Unless $\mu$ is assumed to be $\sigma$-finite, I'm not sure that such a sequence need exist. But it seems to me that the result follows from the Hahn decomposition theorem, with $E^*$ being the union of all positive sets and $E_*$ the union of all negative sets. Oct 10, 2022 at 10:45
• Hm @Math1000. We had yet to even state the Hahn-Jordan decomposition or variation of measures when we proved this. Thanks for the input! Maybe I missed the assumption that $\mu$ is $\sigma$-finite somewhere. Oct 10, 2022 at 14:00

The issue has nothing to do with measure theory. It is just about $$\sup$$ in $$\mathbb{R}$$.
What you need, is to be precise by what you mean by $$\begin{equation*} a = \sup S, \end{equation*}$$ where $$S \subset \mathbb{R}$$. To me, it means that no element in $$S$$ is bigger then $$a$$; and also, that for any $$n \in \mathbb{N}^*$$, there is $$s_n \in S$$ such that $$\begin{equation*} a - \frac{1}{n} \leq s_n \leq a. \end{equation*}$$ But this implies that $$s_n \rightarrow a$$.
In your case, each $$s_n$$ is of the form $$s_n = \mu(E_n)$$ for some $$E_n \in \mathcal{F}$$.