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Math Help - Show some results

  1. #1
    Junior Member
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    Show some results

    Hi!

    I have two questions:

    Let (\Omega, \mathcal{F}, P) be a probability space.
    Now use the properties of the probability measure  P to show

    1) If A\in \mathcal{F}, B\in \mathcal{F} and  A \subset B , then P(A) \leq P(B).

    I tried: If A\subset B, then A+C=B, where A\cap C = \emptyset \Leftrightarrow A+C=A\cup C .
    Then by the countable additivity rule, P(B)=P(A)+P(C) .
    But P(C) \geq 0 , hence P(A) \leq P(B) .

    2) Show that if A\in \mathcal{F} and  \left\{A_{n}\right\}_{n=1}^{\infty} is a sequence of sets in \mathcal{F} with  \lim_{n \to \infty} P(A_{n})=0 and A \subset A_n for all  n , then P(A)=0.

    I tried: A_n=A+C_n, \, A \cap C = \emptyset . P(A_n)=P(A)+P(C_n)
    0=\lim_{n\to \infty} P(A_n) = P(A) + \lim_{n\to \infty} P(C_n)
    This means
    P(A)=- \lim_{n \to \infty} P(C_n)
    But  P(C_n) and  P(A) are greater than zero, so P(A)=0

    Are these valid solutions?

    Thank you
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  2. #2
    MHF Contributor

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    Your proof for #1 is correct.

    Could you use #1 to prove #2?
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  3. #3
    Junior Member
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    Thank you!

    I can use #1 by setting  B = \left\{A_n\right\}_{n=1}^{\infty} ?
    You mean making the solution easier? Is the solution to 2) wrong as it stands now?
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  4. #4
    Senior Member
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    I think the second proof works if you shore up some typos. IMO a more clean way to prove it would be to let \epsilon > 0 be given, find n so that P(A_n) < \epsilon and note that P(A) \le P(A_n).
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  5. #5
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    Thanks to both of you.
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