This is the problem I work on it:-

Let $\left\langle S,\mathcal{F}\right\rangle$ be a measurable space, and suppose that the $\sigma$-algebra $\mathcal{F}$ contains an infinite collection of sets. Prove that for every integer $n\geq 1$, the $\sigma$-algebra $\mathcal{F}$ contains exactly $n$ disjoint nonempty measurable sets.

My attempt:- Since $\mathcal{F}$ contain infinite collection of sets, for any $n\ge 1$ we can take an arbitrary $n$ sets $ E_{1},E_2\dots E_n$ and of $E_{i}\ne E_{j}$ for $i\ne j$. now this is how I make the $n$ disjoint set let us call it $F_n$.

$$F_1=E_1$$ $$F_2=E_2 \backslash E_2\cap E_1$$ $$F_i=E_i \backslash \cup_{k=1}^{i-1} (E_i\cap E_k)$$ so I am kind sure this the family $\{F_k\}_{k=1}^{n}$ is disjoint, but there is possibility that many of them be empty. please help to correct my answer.


For $n=1$ we take $S$.

Assume it is true for $n$ and $E_1,...,E_n$ are disjoint and non-empty. All possible unions of these sets form a finite collection. Since $\mathcal{F}$ is infinite there is $E\in\mathcal{F}$ that is not in this collection. Take

$$E_{n+1}':=E\setminus \cup_{E_i\subset E} E_i,$$

$$E_i':=E_i\setminus E$$

for $i=1,...,n$ with $E_i$ not contained in $E$, and


for $E_i\subset E$.

The new collection $E_1',...,E_{n+1}'$ is disjoint and non-empty.

  • $\begingroup$ I like your solution because you used indication, which I never come to my mind to use it in this problem.thanks $\endgroup$ – henry Jan 2 '15 at 4:12

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.