# Proof that $A\subseteq B\implies\Bbb P(A) \le\Bbb P(B)$

I have to prove that $A\subseteq B\implies\Bbb P(A) \le\Bbb P(B)$, where $\Bbb P$ is probability.

So far I have learned about direct proof, mathematical induction and proof by contraposition.

This exact proposition is solved in my literature (Elementare Einführung in die Wahrscheinlichkeitsrechnung by Karl Bosch), but it isn’t explained at all and I don’t want to copy sth. I don’t understand.

Which method should I choose and how is it done?

Thanks!

(c means “is subset of”)

#### Solutions Collecting From Web of "Proof that $A\subseteq B\implies\Bbb P(A) \le\Bbb P(B)$"

Since $A\subseteq B\rightarrow N(A)\leq N(B)\rightarrow \frac{N(A)}{N(\Omega)}\leq \frac{N(B)}{N(\Omega)}\rightarrow P(A)\leq P(B)$.

Probability is a kind of measure measuring the size of sets, which are called events. And basically, measure requires two properties. One, non-negative; Two, measure of disjoint union of sets equals to sum of measures of each set. Write in language of probability, for any disjoint events $A,B$, we have
$$P(A\cup B)=P(A)+P(B)\geq0$$

So, if $A\subseteq B$, then
$$P(B)=P(A\cup(B/A))=P(A)+P(B/A)\geq P(A)$$