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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.

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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).

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  • $\begingroup$ Ok, but are there measurable sets other than the sets of measure 0? $\endgroup$ – el_tenedor May 22 '15 at 23:06
  • $\begingroup$ i mean ofcourse yes because the whole space has measure 1. $\endgroup$ – J.E.M.S May 22 '15 at 23:08

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