PROOF 161
Prove.
Probability lies between 0 and 1
On a probability space.
A way to read this proof
First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.
Lemma L1
P(∅) = 0
Proof.
(1)S ∪ ∅ = S, and S and ∅ are disjoint.A4
(2)P(S ∪ ∅) = P(S) + P(∅)(1), A3
(3)Since S ∪ ∅ = S, P(S ∪ ∅) = P(S)(1), E
(4)P(S) = P(S) + P(∅)(2), (3), E
(5)P(∅) = 0(4), A6
∎
Lemma L2
For every event A, P(A) + P(Aᶜ) = 1
Proof.
(1)S = A ∪ Aᶜ, and the two events are disjoint.A5
(2)P(S) = P(A) + P(Aᶜ)(1), A3
(3)P(S) = 1A2
(4)P(A) + P(Aᶜ) = 1(2), (3), E
∎
Theorem T1
For every event A, 0 ≤ P(A) ≤ 1
Proof.
(1)P(A) ≥ 0A1
(2)P(A) + P(Aᶜ) = 1L2
(3)P(Aᶜ) ≥ 0A1
(4)P(A) ≤ P(A) + P(Aᶜ)(3), A7
(5)P(A) ≤ 1(2), (4)
(6)0 ≤ P(A) ≤ 1(1), (5)
(7)The upper bound is not a stipulation. The value 1 is shared between A and Aᶜ, and since the share of Aᶜ is non-negative, the share of A cannot exceed 1.(2), (3), (5)
∎
Corollary C1
Both 0 and 1 are probabilities of some event, so this range cannot be narrowed.
Proof.
(1)P(∅) = 0L1
(2)P(S) = 1A2
(3)0 ≤ P(A) ≤ 1, and both endpoints are attained, so no narrower interval contains every event.(1), (2), T1
∎
Remark
Can A1 be removed from this proof?
Non-negativity supplies the lower bound. It seems unrelated to the upper bound. Remove it and see which side disappears.