Let $x \in \Omega$ and $\mathcal{A} \subset\mathcal{P}(X)$ an at most countable set with $\emptyset, \Omega \in \mathcal{A}$. Let

$$ \delta_x: \mathcal{A} \to [0,\infty], \quad A \mapsto \begin{cases} 1 &\text{if } x\in A\\ 0 &\text{else}. \end{cases}$$

Now consider the extension of $\delta_x$ to the outer measure $\overline\delta_x$ on $\mathcal{P}(X)$ by the Caratheodory construction. Which are the sets of measure zero with respect to $\overline\delta_x$ and of which sets does the $\sigma$-algebra of $\overline\delta_x$-measurable sets $\Sigma(\overline\delta_x)$ consist?

The problem here is the generality of $\mathcal{A}$, since for a set $N \in X$ with $x \not\in N$ we don't know a priori that there will be a countable covering $(B_k)_{k \in \mathbb{N}}$ of $N$ with $B_k \in \mathcal{A}$ and $x \not\in B_k$ for all $k \in \mathbb{N}$. But if such a covering exists, $N$ is a set of measure zero.

What about the other $\overline\delta_x$-measurable sets? How would I characterize them? I think I should use the Definition of $\overline\delta_x$-measureability as

A set $A \subset X$ is $\overline\delta_x$-measureable iff for all $T \subset X$ $$ \overline\delta_x(A \cap T) + \overline\delta_x(A^C \cap T) \leq \overline\delta_x(T). $$

I have the suspicion that $\Sigma(\overline\delta_x)$ only consists of the sets of measure zero and their complements, but I'm not sure how to prove this.


1 Answer 1


Actually you have answered what sets are of measure $0$. They are exactely subsets of union of all $B \in \mathcal{A}$ such that $x \not \in B$. Asume that there is mesurable set $A$, than: $$(1) \: \: \: \:\delta_x(A)+\delta_x(A^c)=1$$ and what is more measure of every set is $0$ or $1$ so if $A$ wasn't of measure $0$ it was of measure $1$ and thanks to $(1)$ its complement is of measure $0$ so only mesurable sets are of measure $0$ or are their complements (as you suspected).

  • $\begingroup$ Ok, but are there measurable sets other than the sets of measure 0? $\endgroup$
    – el_tenedor
    May 22, 2015 at 23:06
  • $\begingroup$ i mean ofcourse yes because the whole space has measure 1. $\endgroup$
    – J.E.M.S
    May 22, 2015 at 23:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.