2.13 Complements and Details 223
and, therefore, δ(ε):= sup{| f (x) − f (y)| : |x − y|≤ε}→0asε ↓ 0.
Define
f
ε
(x) = f ∗ ρ
ε
(x):=
"
ε
−ε
f (x − y)ρ
ε
(y) dy, (C11.8)
and note that, since f
ε
(x) is an average over values of f within the
interval (x − ε, x + ε), | f
ε
(x) − f (x)|≤δ(ε) for all x. Hence,
)
)
)
)
"
R
1
fdP
n
−
"
R
1
f
ε
dP
n
)
)
)
)
≤ δ(ε) for all n,
)
)
)
)
"
R
1
fdP−
"
R
1
f
ε
dP
)
)
)
)
≤ δ(ε),
)
)
)
)
"
R
1
fdP
n
−
"
R
1
fdP
)
)
)
)
≤
)
)
)
)
"
R
1
fdP
n
−
"
R
1
f
ε
dP
n
)
)
)
)
+
)
)
)
)
"
R
1
f
ε
dP
n
−
"
R
1
f
ε
dP
)
)
)
)
+
)
)
)
)
"
R
1
fdP−
"
R
1
fdP
)
)
)
)
≤ 2δ(ε) +
)
)
)
)
"
R
1
f
ε
dP
n
−
"
R
1
f
ε
dP
)
)
)
)
→ 2δ(ε)asn →∞.
Since ε>0 is arbitrary and δ(ε) → 0asε → 0, it follows that
!
R
1
fdP
n
→
!
R
1
fdP, as claimed. Next let F
n
, F be the distribution
functions of P
n
, P, respectively (n = 1, 2,...), and suppose (a) holds
and observe that (−∞, x]isaP−continuity set if and only if 0 =
P(∂(−∞, x]) ≡ P({x}). That is, x must be a continuity point of F.
Therefore, that (a) implies (c), follows from Alexandrov’s theorem. To
show that the converse is also true, suppose F
n
(x) → F(x) at all points
of continuity of a distribution function (d.f.) F. Consider a continuous
function f that vanishes outside [a, b] where a, b are points of continuity
of F. Partition [a, b] into a finite number of small subintervals whose
end points are all points of continuity of F, and approximate f by a
step function constant over each subinterval. Then the integral of this
step function with respect to P
n
converges to that with respect to P and,
therefore, one gets (a) by a triangle inequality.
Remark C11.1 It is straightforward to extend the proof of the equiv-
alence of (a), (b), and (c) to
R
k
, k ≥ 1. The equivalence of (c) and (d)
for the case
R
k
, k > 1, is similar, but one needs to use P-continuities
for approximating finite and infinite rectangles. See Billingsley (1968,
pp. 17–18).