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Let $\Sigma _1,\Sigma _2\subset 2^\Omega$ be independent $\sigma$-algebras and suppose they both contain events for which $0<P(X)<1$.

Show that $\Sigma := \Sigma_1\cup\Sigma _2$ cannot be a $\sigma$-algebra.

Assume $\Sigma$ is a $\sigma$-algebra. Let $A_i\in \Sigma _1, j\in I$ be the events for which $0<P(A_i)<1$ and $B_i\in\Sigma _2,i\in J$ events for which $0<P(B_i)<1$.

Due to independence $0<P(A_i\cap B_j)<1$ so $A_i\cap B_j\in\{\emptyset, \Omega\}$ is impossible. Pick $(k,l)\in I\times J$, then $A_k\cap B_l\in \{A_i\}_{i\in I}$ or $A_k\cap B_l\in\{B_i\}_{i\in J}$. If $A_k\cap B_l = A_k$ (or $A_k^c$), then $$P(A_k)P(B_l) = A_k\Longrightarrow P(B_l)=1\quad\mbox{and}\quad A_k\cap B_l = A_k^c\Longrightarrow A_k\cap B_l =\emptyset, $$ which are contradictions.

Question:
What is the problem with $A_k\cap B_l = A_m$ (or $A_m^c$) ?

If we considered a simpler case e.g $$\Sigma _1 = \{\emptyset,\Omega, A,A^c\}\quad\mbox{and}\quad \Sigma _2 = \{\emptyset,\Omega, B,B^c,C,C^c,D,D^c\} $$ How would one exhibit a contradiction for $A\cap B = C$? (Assuming $\Sigma _1$ and $\Sigma _2$ are independent)

We don't, of course, have to only consider intersections, but since there is some talk about independence, then intersection seems the logical place to look for a contradiction.


In an attempt to play around with it: $$A\cap B= C\Longrightarrow A\cap (B\cap C) = C\Longrightarrow P(A)P(B\cap C) = P(C) \Longrightarrow P(B) = P(B\cap C) $$ Likewise $$ A\cap B = C\Longrightarrow A\cap B=B\cap C\Longrightarrow P(A)P(B) = P(B\cap C)$$ so we get $$P(A)P(B) = P(B) \Longrightarrow P(A)=1$$ which seems problematic. Turns out to be sufficient.

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    $\begingroup$ $P(A)=1$ does not mean that $A$ is certain. Also, $P(A)=0$ does not mean that $A$ is impossible. Think of a shape in $R^2$ over which we have a uniform distribution. Choosing randomly any of these point is possible even though their probability is $0$. $\endgroup$
    – zoli
    Oct 14 '17 at 18:17
  • $\begingroup$ @zoli Good catch! I changed the wording, but the result shouldn't change. $\endgroup$
    – AlvinL
    Oct 14 '17 at 18:19
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If $A_k\cap B_l = A_m$, then $$A_k\cap B_l\cap A_m = A_m\Longrightarrow P(B_l)P(A_k\cap A_m) = P(A_m) \Longrightarrow P(B_l)P(A_k\cap A_m) = P(B_l)P(A_k).$$ Additionally, $$A_k\cap B_l = A_m\cap A_k\Longrightarrow P(B_l)P(A_k) = P(A_m\cap A_k). $$ Necessarely $P(A_k\cap A_m)>0$, implying $P(B_l) =1$, a contradiction.

Similar contradiction for $A_k\cap B_l = B_m$.

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