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Caleb Proctor 2022-07-07 Answered
Fix some probability space ( Ω , F , P ), and let A F be a sub- σ-algebra. Suppose that g , h : Ω R are two P -integrable and A -measurable functions.
I need help understanding why { h > g } A ? I understand that since g and h both are A -measurable, it holds that for all B B ( R ):
h 1 ( B ) A , g 1 ( B ) A .
However, I can't seem to make the connection as to why { h > g } A also holds true.
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Answers (1)

Keegan Barry
Answered 2022-07-08 Author has 18 answers
{ h > g } = { h g > 0 } A because if h and g are A -measurable, then the difference h g is also A -measurable.
To see this, note that for all t R , h g < t h < t + g h < q < t + g for some rational q. So
{ h g < t } = q Q { h < q < t + g } = q Q { h < q } { g > q t }
Hence { h g < t } A as a countable union of elements of A , and therefore h g is A -measurable.

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