$\newcommand{\scrF}{\mathcal{F}}$ 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?

  • $\begingroup$ 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. $\endgroup$
    – Math1000
    Oct 10, 2022 at 10:45
  • $\begingroup$ 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. $\endgroup$ Oct 10, 2022 at 14:00

1 Answer 1


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 $E_n \in \mathcal{F}$.


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