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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.

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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

$$E_i':=E_i$$

for $E_i\subset E$.

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

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  • $\begingroup$ 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

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