Miscellaneous 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
Hence, 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)
Hence, 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) .
Hence, Proposition 3 is proven.