1. ## 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

2. Your proof for #1 is correct.

Could you use #1 to prove #2?

3. 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?

4. 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 $\displaystyle P(A) \le P(A_n)$.

5. Thanks to both of you.