APPENDIX D. PROOFS OF SOME THEOREMS

D-1 Classes of events equal with probability 1

Theorem 3-10B

Let and be finite or countably infinite classes such that = [P]. Then the following three conditions must hold:

PROOF Condition 3 follows from condition 3 of Theorem 3-10A. Condition 1 is established by the following calculations:

since P(A_iB_i^c) = 0 for every i (Theorem 3-10.4).

Interchanging the role of A_i and B_i gives the same expression for .

Now

since P[A_iB_j(A_iB_i)^c] = P[A_iB_j(A_i^c ∪ B_i^c)] = 0 for every i, j.

We have thus established the equalities required by the definition.

Condition 2 follows by considering complements, since

The theorem follows from conditions 1 and 3, which are established.

Theorem 3-10C

If a finite or countably infinite class has the properties

then there exists a partition such that = [P].

PROOF Suppose the events of the class are numbered to form a sequence A₁, A₂, …, A_n, …. First, consider the case in which the union of the A_i is the whole space. Put

Then it is easy to see that the class = {B_i: i ∈ J} is a partition, for the sets are disjoint and their union is the whole space. We note, further, that

and

Thus the class is a partition which is equal with probability 1 to the class . If the union of the class is not the whole space, we simply consider the set A₀, which is the complement of this union and whose probability is zero, and take as B₁ the union of A₁ and A₀. It is evident that in this case B₁ = A₁ [P].

D-2 Random variables equal with probability 1

Theorem 3-10D

Consider two simple random variables X(·) and Y(·) with ranges T₁ and T₂, respectively. Let T = T₁ ∪ T₂, t_i ∈ T. Put A_i = {: X() = t_i} and B_i = {: Y() = t_i}. Then X(·) = Y(·) [P] iffi A_i = B_i [P] for each i.

PROOF We note that if t_i is not in T₁, A_i = , and if t_i is not in T₂, B_i = .

1. Given X(·) = Y(·) [P]. To show A_i = B_i [P]. Put D = {: X() ≠ Y()}. P(D) = 0.

Let . Now on D_i we must have X ≠ Y, so that D_i ⊂ D. This implies P(D_i) = 0 and hence that A_i = B_i [P]. The argument holds for any i.

2. If A_i = B_i [P], then P(D_i) = 0. Since , we have = 0, so that X(·) = Y(·) [P].

Theorem 3-10E

Consider two random variables X(·) and Y(·). Put E_M = {: X() ∈ M} = X⁻¹(M) and F_M = {: Y() ∈ M} = Y⁻¹(M) for any Borel set M. Then, (A) X(·) = Y(·) [P] iffi (B) E_M = F_M [P] for each Borel set M.

PROOF

1. To show (A) implies (B) we consider

from which it follows that P(A) = P(AS₁) + P(AS₁^c) = P(AS₁) for every event A. Moreover, a little reflection shows that E_MS₁ = F_MS₁ = E_MF_MS₁ for any Borel set M. We thus obtain the fact that P(E_M) = P(F_M) = P(E_MF_M), which is the defining condition for E_M = F_M [P].

2. To show (B) implies (A), we consider first the case that X(·) and Y(·) are simple functions. Define the sets T₁, T₂, T, A_i, and B_i as in the previous theorem. Then A_i = X⁻¹({t_i}) and B_i = Y⁻¹({t_i}). By hypothesis, A_i = B_i [P], which, by Theorem 3-10D, above, implies X(·) = Y(·) [P].

In the general case, X(·) is the limit of a sequence X_n(·), and Y(·) is the limit of a sequence Y_n(·), n = 1, 2, …. Under hypothesis (B) and the construction of the approximating functions X_n(·) and Y_n(·), X_n⁻¹({t_in}) = X⁻¹[M(i, n)] and Y_n⁻¹({t_in}) = Y⁻¹[M(i, n)] are equal with probability 1. Thus X_n(·) = Y_n(·) [P]. If we put C_n = {: X_n() ≠ Y_n()}, we must have P(C_n) = 0. Let Then P(C) = 0. On the complement C^c we have X_n() = Y_n() for all n, so that the two sequences approach the same limit for each in C^c. This means that X() = Y() for all in C^c. Since P(C^c) = 1, the theorem is proved.

Theorem 3-10F

Suppose X(·) and Y(·) are two nonnegative random variables. Then (A) X(·) = Y(·) [P] iffi (B) there exist nondecreasing sequences of simple random variables {X_n(·): 1 ≤ n < ∞} and {Y_n(·): 1 ≤ n < ∞} satisfying the three conditions:

1. X_n(·) = Y_n(·) [P], 1 ≤ n < ∞

PROOF

1. To show (A) implies (B), we form the simple functions according to the scheme described in Sec. 3-3. Let the points in the subdivision for the nth approximating functions be t_in. Put E(i, n) = {: X_n() = t_in] and E′(i, n) = {: Y_n() = t_in}. By Theorem 3-10E (proved above), we must have E(i, n) = E′(i, n) [P] for each i, n, so that X_n(·) = Y_n(·) [P] for every n. By Theorem 3-3A, properties 2 and 3 must hold.

2. To show (B) implies (A), let D_n = {: X_n() ≠ Y_n()}. Then P(D_n) = 0, so that, if , we must have P(D) = 0. On D^c, X_n() = Y_n() for each , n, so that for each ∈ D^c. Thus X(·) = Y(·) on D^c, so that the set on which these fail to be equal must be a subset of D and thus have probability zero.

D-3 Proof of Theorem 6-1F on the law of large numbers

Theorem 6-1F

Suppose the class of random variables {X_i(·): 1 ≤ i < ∞} is pairwise independent and each random variable in the class has the same distribution with [X_i] = . The variance may or may not exist. Then for any > 0.

PROOF The proof makes use of a classical device known as the method of truncation. Let δ be any positive number. For each pair of positive integers i, n, define the random variables Z_in(·) and W_in(·) as follows: