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Consider the measurable space $(\mathbb{R}, \Sigma)$, where $$\Sigma := \{ A \subset \mathbb{R} \,:\, A \text{ is countable or } A^c \text{ is countable}\}.$$

Proving this is indeed a $\sigma$-algebra is easy: The countable union of countable sets is again countable; if the countable union contains at least one cocountable set, the coset of this union will be a subset of a countable set and thus countable.

Now consider the following map:

$$\mu: \Sigma \to [0,\infty], \quad A \mapsto \begin{cases}0 \quad \text{if $A$ is countable} \\ 1 \quad \text{else}\end{cases}$$

How would I prove that this map is a measure? Again, if I only consider disjoint unions of countable sets $\sigma$-additivity is obvious. But what about disjoint unions that contain uncountable sets with countable complement?

I suspect that every disjoint union of sets in $\Sigma$ contains at most one uncountable set with countable coset, but I can't find a rigorous proof for this. The only uncountable sets $A \in \Sigma$ with countable coset I can picture at this moment are of the form

$$A = \mathbb{R} \setminus Q,$$

where $Q \subset \mathbb{Q}$.

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2 Answers 2

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Let $A=\bigcup_{n=1}^{\infty}A_n$ with $A_n\in\Sigma$ disjoint.

If $A_n$ is countable for each $n$ then also $A$ is countable so that: $$\mu(A)=0=\sum_{n=1}^{\infty}\mu(A_n)$$

This was allready noted by yourself.

Now let it be that e.g. $A_1$ is cocountable.

Then also $A$ is cocountable. Secondly the disjointness of the sets $A_n$ implies that $A_n\subseteq A_1^c$ for $n=2,3,\dots$ so as subsets of a countable set these sets must be countable.

Then:$$\mu(A)=1=1+0+0+\cdots=\sum_{n=1}^{\infty}\mu(A_n)$$

It is obvious that this also works if another set (e.g. $A_{45}$) is taken to be cocountable.

Proved is now that $\mu$ is $\sigma$-additive.

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Assume that there are two disjoint uncountable sets $A, B$ with countable complements. Then $\mathbb{R} = (A\cap B)^c = A^c \cup B^c$ would be the union of two countable sets, in contradiction to the uncountability of the real numbers.

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