An algebra $A$ is a $\sigma$-algebra if and only if $A$ is closed under countable increasing unions.

Proof: Suppose $\{E_j\}_{1}^{\infty}\subset A$ and $E_1\subset E_2\subset \ldots$ Set $$F_k = E_k \setminus \big[\bigcup_{1}^{k-1}E_j\big] = E_k \cap \big[\bigcup_{1}^{k-1}E_j\big]^{c}$$ Then the $F_k$\s belong to $A$ and are disjoint, and $$\bigcup_{1}^{\infty}E_j = \bigcup_{1}^{\infty}F_k$$ Therefore, $\bigcup_{1}^{\infty}E_j\in A$

I am not sure if this is correct, any suggestions is greatly appreciated


Clearly a $\sigma$-algebra, $\mathscr{A}$ on a set $X$ is by definition closed under "all kinds of" countable unions of sets that belongs to $\mathscr{A}$.

Now suppose $\mathscr{A}$ is an algebra on $X$, which is closed under countable increasing unions.

Pick an arbitrary sequence of sets $\left\{A_{n}\right\}_{n=1}^{\infty}\subset \mathscr{A}$ and consider the sequence $\left\{B_{k}\right\}_{k=1}^{\infty}$, defined by $$B_{k}= \bigcup_{n=1}^{k}A_{n}$$

Now indeed $B_{k}\subseteq B_{k+1}$ and $B_{k}\in \mathscr{A}$, for all $k\geq1$, since $\mathscr{A}$ is an algebra. Moreover $$\bigcup_{n=1}^{\infty}A_{n}=\bigcup_{k=1}^{\infty}B_{k}\in \mathscr{A}$$

which finishes the proof.

  • $\begingroup$ Looks good, so is my proof wrong or was there something missing? $\endgroup$ – Wolfy Aug 12 '15 at 19:46
  • $\begingroup$ @Wolfy For what it's worth, yes, your proof too, is correct. $\endgroup$ – Erfan Oct 6 '17 at 7:53

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