Prove Theorem: Probability of A Subset B is Less than B

[Guest2]

How do I prove the following theorem?

If $A \subset B$ then $P(A) \le P(B)$ and $P(B-A) = P(B)-P(A)$

$A$ and $B$ are events and $P$ is the probability function.

What I tried (but not sure if it's right or not):

$P(B) = P((B\setminus A) \cup (B \cap A)) = P(B\setminus A)+P(B \cap A) \ge 0+P(B \cap A) \ge P(A) $

Therefore $P(B) - P(A) \ge 0$. Hence $P(B) \ge P(A)$.

 

[I like Serena]

How do I prove the following theorem?

If $A \subset B$ then $P(A) \le P(B)$ and $P(B-A) = P(B)-P(A)$

$A$ and $B$ are events and $P$ is the probability function.

What I tried (but not sure if it's right or not):

$P(B) = P((B\setminus A) \cup (B \cap A)) = P(B\setminus A)+P(B \cap A) \ge 0+P(B \cap A) \ge P(A) $

Therefore $P(B) - P(A) \ge 0$. Hence $P(B) \ge P(A)$.

Hi Guest! ;)

It looks right to me.
You may want to list the axioms you're using though.
There is some context missing, which is different in every textbook on the subject.

 

[Guest2]

Hi Guest! ;)

It looks right to me.
You may want to list the axioms you're using though.
There is some context missing, which is different in every textbook on the subject.

Hi, I like Serena. :D Thanks for the reply.

I'm using the following three axioms:

Axiom 1: For every event $A$ in the class $C$,

$$P(A) \ge 0$$

Axiom 2: For the sure or certain event $S$ in the class $C$,

$$P(S) = 1$$

Axiom 3: For any number of mutually exclusive events $A_1, A_2,\cdots$, in the class C,

$$P(A_1 \cup A_2 \cup \cdots) = P(A_1)+P(A_2) +\cdots $$

How do I prove the second part of the theorem?

I know that $P(B-A) = P(B \cap A')$ but I don't know how to turn the intersection into union so that I can use axiom 3.

 

[I like Serena]

How about $B=(B-A)\cup A$?