Let $\Omega=\{1,3,5,7,9\}$ and $\mathcal{U}=\{\{1,3,5\},\{1,5,7,9\}\}$. I am trying to find the smallest $\sigma$-algebra, $\sigma_\mathcal{U}$, generated by $\mathcal{U}$.

By considering the partition $\{3\}, \{1,5\}, \{7,9\}$, I found that $$\sigma_\mathcal{U}=\{\emptyset,\{1,3,5,7,9\},\{3\},\{1,5\},\{7,9\},\{1,3,5\},\{1,5,7,9\},\{3,7,9\}\}.$$ I know that the $\sigma_\mathcal{U}$ generated by $\mathcal{U}$ is the power set of my partition and should therefore contain eight elements.

Is my logic correct? I'm somewhat of a novice at measure theory.


1 Answer 1


Your logic is indeed correct. A $\sigma$-algebra containing $\mathcal U$ would have to contain $\mathcal V = \{\{3\}, \{1,5\}, \{7,9\}\}$ and hence the power set $\mathcal P(\mathcal V)$.

As the smallest $\sigma$-algebra containing $\mathcal U$ is the intersect of all $\sigma$-algebras containing $\mathcal U$, it is equal to $ \sigma_{\mathcal U}$.


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