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  1. #1
    Junior Member
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    Show some results

    Hi!

    I have two questions:

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

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

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

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

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

    Are these valid solutions?

    Thank you
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  2. #2
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    Your proof for #1 is correct.

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

    I can use #1 by setting $\displaystyle 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
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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 $\displaystyle \epsilon > 0$ be given, find $\displaystyle n$ so that $\displaystyle P(A_n) < \epsilon$ and note that $\displaystyle P(A) \le P(A_n)$.
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  5. #5
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    Thanks to both of you.
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