1. ## 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

2. Your proof for #1 is correct.

Could you use #1 to prove #2?

3. 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?

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 $\epsilon > 0$ be given, find $n$ so that $P(A_n) < \epsilon$ and note that $P(A) \le P(A_n)$.

5. Thanks to both of you.