i was wondering whether it makes any difference (or whether it is even true) that an uncountably infinite union/intersection of sets that are elements of a sigma algebra is again an element of the sigma algebra?

  • 2
    $\begingroup$ Generally, an uncountable union or intersection will not belong to the $\sigma$-algebra. $\endgroup$ – Daniel Fischer Oct 21 '13 at 14:54
  • $\begingroup$ For instance, assume that the Borel sigma algebra on $\mathbb{R}$ is closed under uncountable unions. Then any subset $A\subseteq\mathbb{R}$ would be Borel, since $A=\bigcup_{x\in A} \{x\}$ and since $\{x\}$ is Borel for all $x$. This is clearly not the case. $\endgroup$ – Stefan Hansen Oct 21 '13 at 15:04
  • $\begingroup$ Also see this : math.stackexchange.com/questions/423883/… $\endgroup$ – Prahlad Vaidyanathan Oct 21 '13 at 15:05
  • $\begingroup$ When you have the time look up the "suslin operation". It is a particular instance of an uncountable union which preserves measurability. $\endgroup$ – Umberto P. Oct 21 '13 at 15:50

If $X$ is a set and $\mathcal B\subset 2^X$ a $\sigma$-algebra on $X$, it's not true. For example, take $\mathbb R$ endowed with the $\sigma$-algebra $\mathcal B$ of the subset of $\mathbb R$ which are at most countable, and their complements. Then write $(0,\infty):=\bigcup_{x\gt 0}\{x\}$.


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