0
$\begingroup$

I am reading Schilling's “Measures, integrals and martingales”, where on page 15 he constructs a $σ$-algebra, according to: $$ \mathcal{A} = \lbrace A \subset X: \# A \leq \# \mathbb{N} \quad \text{or} \quad \# A^c \leq \# \mathbb{N} \rbrace, $$ i.e. those sets which are countable/have cardinality smaller than or equal to the natural numbers.

He then goes on to show that this set fulfills the properties of $σ$-algebras; the one I am struggling with is:

  • $A \in \mathcal{A} \implies A^c \in \mathcal{A}$.

The author's argument is:

  • If $A\in\mathcal{A}$, either $A$ or $A^c$ is by definition countable, so $A^c \in \mathcal{A}$.

I am not sure I can agree with this being so simple: if $A$ is indeed countable (so we choose the $\#A \leq \#\mathbb{N}$ condition), then it is already part of the set; its complement $A^c$ need not be countable as per the definition.

I am not sure I understand how the author relates $A$ being countable to its complement being countable? In fact, we don't know anything about its complement, other than it being the complement with respect to $X$ (which we also don't know if it is countable or not).

$\endgroup$
  • $\begingroup$ How do we know that all countable unions are in the σ-algebra? $\endgroup$ – Jos van Nieuwman Jun 1 '19 at 17:15
2
$\begingroup$

$(A^c)^c = A$. If $A$ is countable then $(A^c)^c$ is countable, and if $A^c$ is countable then $A^c$ is countable. Either way, $A^c$ or its complement is countable.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Priceless; thank you! :) $\endgroup$ – mSSM Apr 21 '15 at 23:02

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.