Math 418/544 Proof of Thm. 3.3
We only need to prove d)⇒e). Assume ∀A ∈ B(S) P (∂A) = 0 implies Pn (A) → P (A).
Let f : S → R be bounded Borel s.t. P (Df ) = 0. We must show
Z
Z
lim
f dPn = f dP.
n→∞
(1)
Lemma. If f : S → R and A ⊂ R, then
(a) f −1 (A) ⊂ Df ∪ f −1 (A)
(b) ∂(f −1 (A)) ⊂ Df ∪ f −1 (∂A).
Proof. (a) Let x ∈ f −1 (A) ∩ Dfc . Then there is a sequence {xn } in f −1 (A) s.t. xn → x. As x
is a continuity point of f we have f (xn ) → f (x), and since f (xn ) ∈ A, it follows that f (x) ∈ A.
This proves x ∈ f −1 (A), and (a) follows.
(b) ∂f −1 (A) = f −1 (A) ∩ f −1 (A)c = f −1 (A) ∩ f −1 (Ac ). Now use (a) on each of these sets to
conclude that
∂f −1 (A) ⊂ (Df ∪ f −1 (A)) ∩ (Df ∪ f −1 (Ac )) = Df ∪ (f −1 (A ∩ Ac )) = Df ∪ f −1 (∂A).
For each natural number k choose real numbers xk1 < · · · < xkNk so that
(A) Range(f ) ⊂ [xk1 , xkNk ), (B) xki+1 − xki < 2−k for all i, and (C) P (f −1 ({xki })) = 0 for all i.
The latter condition is easy to satisfy since we only need to avoid the countable set of points for
which P (f −1 ({x})) > 0. Let Aki = f −1 ([xki , xki+1 )) for i = 1, . . . , Nk − 1 and define
PNk −1 k
fk = i=1
xi 1Aki . Then conditions (A) and (B) imply that kf − fk k∞ ≤ 2−k . We also have
by (b) of the Lemma and our assumption that P (Df ) = 0,
P (∂Aki ) = P (∂f −1 ([xki , xki+1 )) ≤ P (Df ∪ f −1 ({xki , xki+1 }) = 0,
the last by condition (C). It follows by hypothesis that limn→∞ P (Aki ) = P (Aki ) for all i, k and
hence that for all k,
Z
Z
NX
k −1
k
lim
fk dPn = lim
Pn (Ai ) = fk dP.
n→∞
n→∞
i=1
The uniform convergence of fk to f therefore shows that for any fixed k,
Z
Z
Z
Z
Z
i
hZ
lim sup f dPn − f dP ≤ lim sup f − fk dPn + fk dPn − fk dP + fk − f dP n→∞
n→∞
−k
≤2
+ 0 + 2−k = 21−k .
As k can be taken arbitrarily large, (1) follows.