1902 CHAPTER 59. BASIC PROBABILITY

are Borel measurable functions. Furthermore, the above functions are

B (E)×·· ·×B (E)

measurable where the above denotes the product measurable sets as described in Theorem12.12.6 on Page 332.

Proof: First consider the case where A =U, an open set. Let

y ∈

{x ∈ Em : µ

(U +

m

∑i=1

xi

)> α

}(59.14.22)

Then from Lemma 59.1.9 on Page 1859 there exists a compact set, K ⊆U +∑mi=1 yi such

that µ (K)> α. Then if y′ is close enough to y, it follows K ⊆U +∑mi=1 y′i also. Therefore,

for all y′ close enough to y,

µ

(U +

m

∑i=1

y′i

)≥ µ (K)> α.

In other words the set described in 59.14.22 is an open set and so y→ µ (U +∑mi=1 yi) is

Borel measurable whenever U is an open set in E.Define a π system, K to consist of all open sets in E. Then define G as{

A ∈ σ (K ) = B (E) : y→ µ

(A+

m

∑i=1

yi

)is Borel measurable

}

I just showed G ⊇K . Now suppose A ∈ G . Then

µ

(AC +

m

∑i=1

yi

)= µ (E)−µ

(A+

m

∑i=1

yi

)

and so AC ∈ G whenever A ∈ G . Next suppose {Ai} is a sequence of disjoint sets of G .Then

µ

((∪∞

i=1Ai)+m

∑j=1

y j

)= µ

(∪∞

i=1

(Ai +

m

∑j=1

y j

))

=∞

∑i=1

µ

(Ai +

m

∑j=1

y j

)

and so ∪∞i=1Ai ∈ G because the above is the sum of Borel measurable functions. By the

lemma on π systems, Lemma 12.12.3 on Page 329, it follows G = σ (K ) = B (E) . Sim-ilarly, x→ µ

(A−∑

mj=1 x j

)is also Borel measurable whenever A ∈B (E). Finally note

thatB (E)×·· ·×B (E)