Let F and G be two sigma-fields on Ω. Prove that F ∩ G is also a sigma-field on Ω.
Show by example that F ∪ G may fail to be sigma-field if Ω = {1, 2, 3}.
"\\emptyset \\in F,\\emptyset \\in G\\Rightarrow \\emptyset \\in F\\cap G\\\\A\\in F\\cap G\\Rightarrow A\\in F,A\\in G\\Rightarrow \\bar{A}\\in F,\\bar{A}\\in G\\Rightarrow \\bar{A}\\in F\\cap G\\\\A_i\\in F\\cap G\\Rightarrow A_i\\in F,A_i\\in G\\Rightarrow \\bigcup{A_i}\\in F,\\bigcup{A_i}\\in G\\Rightarrow \\bigcup{A_i}\\in F\\cap G"
Thus "F\\cap G" is a sigma-algebra.
Let
"F=\\left\\{ \\emptyset ,\\left\\{ 1 \\right\\} ,\\left\\{ 2,3 \\right\\} ,\\left\\{ 1,2,3 \\right\\} \\right\\} \\\\G=\\left\\{ \\emptyset ,\\left\\{ 2 \\right\\} ,\\left\\{ 1,3 \\right\\} ,\\left\\{ 1,2,3 \\right\\} \\right\\} \\\\F\\cup G=\\left\\{ \\emptyset ,\\left\\{ 1 \\right\\} ,\\left\\{ 2 \\right\\} ,\\left\\{ 1,3 \\right\\} ,\\left\\{ 2,3 \\right\\} ,\\left\\{ 1,2,3 \\right\\} \\right\\}"
We see that "\\left\\{ 1 \\right\\} \\in F\\cup G,\\left\\{ 2 \\right\\} \\in F\\cup G,\\left\\{ 1 \\right\\} \\cup \\left\\{ 2 \\right\\} =\\left\\{ 1,2 \\right\\} \\notin F\\cup G"
Thus "F\\cup G" is not a sigma-algebra.
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