Proofs

Measure and Probability (Proofs)

Proposition 1. The measure of the empty set is always 0. In other words:

$m(\emptyset) = 0$

Proof of Proposition 1. Since

$\emptyset \cap \emptyset = \emptyset$

the empty set is disjoint from itself. So, rule 2 of measures implies:

$m(\emptyset \cup \emptyset) = m(\emptyset) + m(\emptyset) \qquad (Eq. \ 1)$

On the other hand,

$\emptyset = \emptyset \cup \emptyset$

implies:

$m( \emptyset ) = m( \emptyset \cup \emptyset ) \qquad (Eq. \ 2)$

Combining Equations (1) and (2) we get:

$m( \emptyset ) = m( \emptyset ) + m( \emptyset ) \qquad (Eq. \ 3)$

Subtract $m(\emptyset)$ from both sides of Equation (3) to get:

$m(\emptyset) = 0$

So Proposition 1 is proven.

Proposition 2. Suppose that A and B are subsets of S and that A is contained in B then

$m(A) \leq m(B)$

Proof of Proposition 2. Write B as the disjoint union:

$B = A \cup (B - A)$

Rule 2 of measures then implies:

$m(B) = m(A) + m(B-A)$

Rule 1 of measures implies:

$m(B-A) \geq 0$

So:

$m(B) \geq m(A)$

So Proposition 2 is proven.

Proposition 3. If $A \subset S$ , then $0 \leq m(A) \leq m(S)$ .

Proof of Proposition 3. A and S are subsets of S, so Proposition 2 implies $m(A) \leq m(S)$ . Rule 1 of measures indicates that $0 \leq m(A)$ .

Professor McCarthy Statistics