# generated sigma-algebras in measure theory

Let $X$ be a set and $E$ and $F$ are classes of subsets on X. then $E \cup F = \{ D \subseteq X : D \in E \text{ or } D \in F\}$.

1. Show that by example $\sigma [E \cup F]$ is not necessarily equal to $\sigma[E] \cup \sigma[F]$
2. Prove that $\sigma [E \cup F] = \sigma[\sigma[E] \cup \sigma [F]]$
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this is my home work and i could not find the solution can any one help me –  math Oct 24 '12 at 15:12
@did: I have the impression OP posted a literal quote from his problem source. Of course, then, the statement should still be indicated as such properly by angle brackets. –  Lord_Farin Oct 24 '12 at 15:20
can u help me i m running late for this problem –  math Oct 24 '12 at 15:22
madhav: Are you going to mention to your tutor the help you received here? –  Did Oct 24 '12 at 15:50

We know that

• $A\subseteq B$ implies $\sigma(A)\subseteq\sigma(B)$
• $\sigma(\sigma(A))=\sigma(A)$.

(These two properties are easy to prove - if you have not seen them in your lecture, you should try to find a proof.)

What do you get if you apply this to $E\cup F\subseteq \sigma(E)\cup\sigma(F)$? What do you get if you apply this to $\sigma(E)\cup\sigma(F) \subseteq \sigma(E\cup F)$? Can you show $\sigma(E)\cup\sigma(F) \subseteq \sigma(E\cup F)$ using these two properties?

For the first part (counterexample) try to think what happens if you divide $X$ into two disjoint parts and you choose sets in $E$ from one of them and sets in $F$ from the other one.

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@Lord_Farin Thanks for correcting the typo. When I saw your username, I wondered whether you're in some relation to Farin Urlaub. –  Martin Sleziak Oct 24 '12 at 15:30
A fresh pair of eyes spots those more easily :). I'm sure you'll return the favour. –  Lord_Farin Oct 24 '12 at 15:33
Nope, not in the least. It's an inheritance from more innocent years; I've simply stuck with it as my preferred internet pseudonym. –  Lord_Farin Oct 24 '12 at 15:37