Dr. Neal, WKU
MATH 532
Sequences of Functions
Throughout, let ( X, F , ) be a measure space and let { f n }∞
n =1 be a sequence of realvalued functions defined on X . We now shall study some properties of this sequence
of functions including various types of convergence.
Theorem 5.1. Assume { f n } are all measurable functions. Let f : X → [−∞, ∞ ] and
g : X → [−∞, ∞] be defined by
f ( x) = sup f n (x )
n
and
g(x ) = inf f n ( x) .
n
Then f and g are measurable.
Proof. Fix x ∈ X . If { f n (x )}∞
n =1 is bounded above (respectively, bounded below), then
f ( x) = sup f n (x ) (respectively, g(x ) = inf f n ( x) ) exists as a finite number. Otherwise,
n
n
we let f ( x) = +∞ (respectively, g(x ) = −∞ ) .
Now let b ∈ℜ . Because sup f n ( x) ≤ b if and only if f n (x ) ≤ b for all n , we have
n
∞
{x : f (x ) ≤ b} = I {x : f n (x ) ≤ b} ∈ F . Thus, f is measurable.
n=1
∞
Also, inf f n ( x) ≥ b if and only if f n (x ) ≥ b for all n . So, {g ≥ b} = I { f n ≥ b} ∈ F .
n =1
n
c
Thus, {g < b} = {g ≥ b}
Exercise 4.1 (a)).
∈ F for all b ∈ℜ , which implies that g is measurable (see
QED
Two Modes of Convergence
Again let { f n }∞
n =1 be a sequence of real-valued functions defined on X and let E ⊆ X .
For each x ∈ X , { f n (x )}∞
n =1 is a real-valued sequence that may or may not converge
with respect to the absolute value metric of the real line.
I. Pointwise Convergence – A sequence of functions { f n } converges pointwise on E
provided { f n (x )}∞
n =1 converges in ℜ for every x ∈ E . Then f ( x) = lim f n (x ) is a welln→∞
defined function on E because limits are unique. We then write f n → f on E .
Then given a fixed x ∈ E we can say: For every > 0, there exists an integer N x ≥ 1
(that depends on the x being used) such that if n ≥ N x then f n ( x) − f ( x) < .
II. Uniform Convergence – A sequence of functions { f n } converges uniformly on E to
the function f ( x) provided: For every > 0, there exists an single integer N ≥ 1 (that
works for every x ∈ E ) such that if n ≥ N then f n ( x) − f ( x) < for all x ∈ E . In this
case, we again write f ( x) = lim f n (x ), but also write f n ⇒ f on E .
n→∞
Dr. Neal, WKU
Theorem 5.2. (a) Uniform convergence always implies pointwise convergence.
(b) If { f n } are measurable and f n → f on X , then f is measurable.
Proof. (a) Follows directly from the definitions of convergence.
(b) Let b ∈ℜ . We shall show that {x : f (x ) ≤ b} is a measurable set. To do so, we
assert that
{x : f (x ) ≤ b} =
I
∞
U
∞
I {x : f n (x ) < b + } ,
∈Q+ N =1 n = N
which is a measurable set because it consists of denumerable intersections/unions of
measurable sets.
Fix x ∈ X . If f ( x) ≤ b , then for every
f n (x ) < f (x ) +
≤b+
> 0 there exists an integer N ≥ 1 such that
for all n ≥ N . Thus, x ∈ I
On the other hand, suppose x ∈
∞
U
∞
I {x : f n (x ) < b + } .
∈Q+ N =1 n= N
∞
∞
I U I {x : f n (x )
∈Q+ N =1 n= N
< b + }.
Let
′ > 0 be
given. Then there exists an ∈Q+ , with 0 < ≤ ′ . And for this ∈Q+ , there is an
integer N ≥ 1 such that f n (x ) < b +
for all n ≥ N .
But this means that
f ( x) = lim f n (x ) ≤ b + ≤ b + ′ for all ′ > 0 . Because f ( x) ≤ b + ′ for all ′ > 0 , we
n→∞
must have f ( x) ≤ b .
QED
1
Example 5.1. Let f n :(0, ∞) → ℜ be defined by f n (x ) = 10 +
for n ≥ 1. We assert
nx
that { f n } converges pointwise on (0, ∞ ) to the function f ( x) = 10, but does not
converge uniformly.
So let x ∈(0, ∞) and let > 0 be given. Let N be the smallest integer such that
1 / ( x ) < N so that 1 / N < x . Then for n ≥ N , we have
f n ( x) − f ( x) = 10 +
1
1
1
x
− 10 =
≤
<
= .
nx
nx N x
x
Thus, f n (x ) → f (x ) (pointwise) for all x ∈(0, ∞). But as we can see, the choice of N
depends on the specific x in the domain.
1
But suppose f n ( x) − f ( x) =
< for all x ∈(0, ∞) if n ≥ N0 for some N0 ≥ 1.
nx
1
Then we would have 0 <
< x for all x ∈(0, ∞), which is a contradiction because we
N0
1
can always find a real number between 0 and
.
N0
Dr. Neal, WKU
Using properties of real-valued sequences, we obtain the following results:
Proposition 5.1. Let f n → f on E and let gn → g on F . Then
(a) For all c ∈ ℜ , c f n → c f on E .
(b) f n + gn → f + g on E ∩ F .
(c) f n gn → f g on E ∩ F .
(d) f n (x )/ gn (x ) → f (x ) / g(x ) for x ∈ E ∩ F (provided g(x ) ≠ 0 and gn (x ) ≠ 0 for all
but a finite number of n ).
Properties (a) and (b) also hold for uniform convergence, but Property (c) does not
hold for uniform convergence without additional boundedness conditions.
We next prove a Cauchy criterion that is equivalent to uniform convergence.
Theorem 5.3. A sequence of functions { f n } converges uniformly on E if and only if for
every > 0 there exists an integer N ≥ 1 such that f n ( x) − f m ( x) < for all x ∈ E
whenever n,m ≥ N .
Proof. Suppose f n ⇒ f . Let > 0. Then there exists an integer N ≥ 1 such that
f n ( x) − f ( x) < / 2 for all x ∈ E whenever n ≥ N . Then for n,m ≥ N we have for all
x ∈ E that f n ( x) − f m ( x) ≤ f n (x ) − f (x ) + f ( x) − f m (x ) < / 2 + / 2 = .
Conversely, suppose the Cauchy criterion holds. Then for each x ∈ E , { f n (x )} is a
Cauchy sequence and therefore has a real-valued limit f ( x) because ℜ is complete.
We assert that f n ⇒ f on E . Let > 0. Then there exists an integer N ≥ 1 such that
f n ( x) − f m ( x) < / 2 for all x ∈ E whenever n,m ≥ N . Let n ≥ N and let x ∈ E . For
this x , we can choose an integer M ≥ N such that
f n ( x) − f ( x) ≤ f n (x ) − f M (x ) + f M (x ) − f ( x) <
/2+
f M ( x) − f (x ) <
/ 2.
/2= .
Then
QED
It is also the case that the uniform limit of continuous functions is continuous.
Theorem 5.4. Let ( X, d X ) be a metric space and let E ⊆ X . Let f n : E → ℜ be a
continuous function for all n ≥ 1, and suppose that f n ⇒ f on E . Then f is continuous
on E .
Proof. Let a ∈ E and let > 0. Then there exists an integer N ≥ 1 such that if n ≥ N then
f n ( x) − f ( x) < / 3 for all x ∈ E . Because f N is continuous at a , there exists > 0
such that if x ∈ E and d X ( x, a) < , then f N (x ) − f N (a) < / 3. So let x ∈ E with
d X ( x, a) < . Then,
f (x ) − f (a) ≤ f (x ) − f N ( x) + f N ( x) − f (a)
≤ f (x ) − f N ( x) + f N ( x) − f N (a) + f N (a) − f (a)
< / 3+ / 3+ / 3 = .
Thus, f is continuous at a .
QED
Dr. Neal, WKU
Theorem 5.4 gives an interesting interchange of limits result in the case when a is also
a limit point of E . Because f is continuous at a and f n (x ) → f (x ) for all x , we have
f (a) = lim f ( x) = lim
x →a
lim f n ( x) .
x → a n →∞
Then because each f n is continuous at a we have,
lim lim f n (x ) = lim f n (a) = f (a) .
n →∞ x → a
n→ ∞
Hence, lim lim f n (x ) = lim lim f n (x ).
n →∞ x → a
x → a n→∞
The next result gives conditions for when pointwise convergence implies uniform
convergence.
Theorem 5.5. Let f n :[ a, b] → ℜ be a sequence of real-valued functions defined on a
closed, bounded interval [a,b] ⊆ ℜ . Assume:
(a) Each f n is continuous on [a,b];
(b) f n → f (pointwise) on [a,b] and that f is also continuous on [a,b];
(c) For each x ∈[a, b], f n (x ) ≥ f n+1 ( x) for all n .
Then f n converges uniformly to f on [a,b].
Proof. Let gn = f n − f for all n . Because f 1 ≥ f 2 ≥ . .. ≥ f n ≥ f n+1 ≥ . . . , we must have
that f n (x ) ≥ lim f n (x ) = f ( x) for all x ∈[a, b]. Thus, gn ≥ 0 for all n . But we also have
n→∞
gn ≥ gn+1 for all n because of the decreasing assumption of { f n }. Because f and each
f n are continuous on [a,b], then gn is also continuous on [a,b] for all n , and we also
have that gn (x ) = f n (x ) − f (x ) → 0 for all x ∈[a, b].
Let > 0 be given. For each n , let Kn = {x ∈[a,b]: gn (x ) ≥ } . If gn+1 (x ) ≥ , then
gn (x ) ≥ also because gn ≥ gn+1 ; thus, Kn+1 ⊆ Kn .


Now suppose Kn ≠ ∅ for all n . Then Kn is actually the set Kn = gn−1  [ , max gn ] ,
 a≤ x≤ b

which is the inverse image of a closed set under a continuous function and therefore
must be closed. But Kn is also a subset of [a,b], so Kn is a closed, bounded set and
therefore must have the form [an , bn ]. So {Kn } is a nested, decreasing set of nonempty intervals and therefore I Kn must contain a point x0 by the Nested Interval
n
Theorem.
However, gn (x 0 ) → 0; thus, there exists an integer N ≥ 1 such that gn (x 0 ) < for
all n ≥ N . That is, x0 ∉ Kn for n ≥ N , which contradicts that x0 ∈ I Kn . So we must
n
have that KM = ∅ for some M , which makes Kn = ∅ for all n ≥ M . Thus, for all n ≥ M
we have f n ( x) − f ( x) = gn (x ) < for all x ∈[a, b]. Hence, f n ⇒ f on [a,b].
QED
Dr. Neal, WKU
Note: The functions f n (x ) = 10 + 1/ (n x) are all continuous on (0, ∞), they satisfy
f n+1 ≤ f n for all n , and they converge pointwise to the continuous function f ( x) = 10.
However, they do not converge uniformly, as shown in Example 4.1. This example
illustrates why we need the additional condition of working on a closed, bounded
interval [a, b].
Definition 5.1. (a) A function f : X → ℜ is bounded if there exists a positive real
number M such that f (x ) ≤ M for all x ∈ X .
(b) A sequence of real-valued functions { f n } is uniformly bounded if there exists a
positive real number M such that for all n , f n ( x) ≤ M for all x ∈ X .
Exercise 5.1. Let { f n }∞
n =1 be a sequence of real-valued functions defined on a set X .
Assume that each f n is a bounded function. Prove:
(a) If { f n } converges uniformly to f on X , then f is bounded.
(b) If { f n } converges uniformly on X , then { f n } is uniformly bounded.
Exercise 5.2. Let { f n } and {gn } be sequences of real-valued functions defined on X that
converge uniformly on E ⊆ X .
(a) Prove that { f n + gn} converges uniformly on E .
(b) Assume also that { f n } and {gn } are sequences of bounded functions. Prove that
{ f n gn } converges uniformly on E .
Exercise 5.3. Construct sequences { f n } and {gn } that converge uniformly on some
subset of ℜ , but such that { f n gn } does not converge uniformly.
Exercise 5.4. Let ( X, d X ) be a metric space and let E ⊆ X . Let f n : E → ℜ be a
continuous function for all n ≥ 1, and suppose that f n ⇒ f on E . For a ∈ E , prove that
lim f n (x n ) = f (a)
n →∞
for every sequence {x n} from E such that xn → a in X .
More Modes of Convergence
The definitions of pointwise and uniform convergence do not require any conditions on
the space X . However the next two forms of convergence do require that we are
working on a measure space and that all functions are assumed to be measurable.
Dr. Neal, WKU
III. Convergence Almost Everywhere – Let ( X, F , ) be a measure space and let { f n }
be a sequence of measurable functions. Then { f n } converges almost everywhere to the
measurable function f provided { f n (x )}∞
n =1 converges in ℜ to f ( x) for every
x ∈ X − A , where A is a set of measure 0. In this case, we write f n → f a.e.
IV. Convergence in Measure – Let ( X, F , ) be a measure space. A sequence of
functions { f n } converges in measure to the measurable function f provided for every
> 0 we have lim ( f n − f ≥ ) = 0 . In this case we write, f n → f .
n →∞
Notes: (a) If f n → f , then f n → f a.e. using the set A = ∅.
(b) On a probability space, we use the alternate phrase converges almost surely and
write f n → f a.s. in place of f n → f a.e.
(c) If f n → f a.e., then
({x : f n (x ) →
/ f (x )}) = 0 .
(d) f n → f if and only if for every
such that
> 0 and every ′ > 0 there exists an integer N ≥ 1
( f n − f ≥ ) < ′ for all n ≥ N .
(e) If { f n } converges in measure to f in a probability space, then we say instead that
{ f n } converges in probability to f and write f n → f .
P
Example 5.2. (a) For n ≥ 1, let f n :[0, ∞ ) → ℜ be defined by
 n

x
f n (x ) =  n + 1
 n
if x ∉ℵ
if x ∈ℵ.
For instance, f 1 (x ) = x / 2 provided x ∉ℵ, but f 1 (x ) = 1 for all x ∈ℵ. Likewise,
f 20 (x ) = 20x / 21 for x ∉ℵ, but f 20 (x ) = 20 for all x ∈ℵ. Then for all x ∉ℵ, we have
n
lim f n ( x) = lim
x = x . But for all x ∈ℵ, lim f n ( x) = lim n = +∞ .
n →∞
n→∞ n + 1
n →∞
n→∞
˜
Now define f : [0, ∞) → ℜ and f : [0, ∞) → [0, ∞] by
f ( x) = x
and
 x if x ∉ℵ
f˜ ( x) = 
+∞ if x ∈ℵ.
Then f n → f˜ everywhere on [0, ∞) . However, with respect to Lebesgue measure
f n → f a.e. on [0, ∞) . That is, f n → f for all x ∈ [0, ∞) −ℵ, and (ℵ) = 0 .
,
Dr. Neal, WKU
(b) For n ≥ 1, let f n :[0, ∞ ) → ℜ be defined by
 10
f n (x ) = 
 0
if x ≥ 1/ n
if 0 ≤ x < 1/ n ,
and let f : [0, ∞) → ℜ be the constant function f ( x) = 10. For every 0 <
on which f n and f differ by at least is always
≤ 10 , the set
{x ∈[0, ∞ ): f n (x ) − f ( x) ≥ } = [0, 1 / n) .
And with regard to Lebesgue measure, lim
n →∞
([0, 1/ n)) = lim 1/ n = 0 . (For
n→ ∞
> 10,
{ f n − f ≥ } = ∅ and (∅) = 0.) Thus, f n → f . It is also the case that f n → f a.e.
for lim f n ( x) = 10 for all x ∈(0, ∞).
n →∞
Proposition 5.2. Let f n → f a.e. and let gn → g a.e. Then
(a)
(c)
(d)
For all c ∈ ℜ , c f n → c f a.e.
(b) f n + gn → f + g a.e.
f n gn → f g a.e.
f n (x )/ gn (x ) → f (x ) / g(x ) a.e. on the set {g ≠ 0}.
The above proposition follows immediately from the properties of real-valued
sequences, with convergence occurring on the set
({x : f n (x ) →
/ f (x )} ∪ {x : gn ( x) →
/ g(x )})c .
The following results show that the limiting functions with regard to convergence
almost everywhere and convergence in measure are unique up to a set of measure 0.
Theorem 5.6. Let ( X, F , ) be a measure space.
(a) If f n → f a.e. and f n → g a.e., then f = g a.e.
(b) If f n → f and f n → g , then f = g a.e.
(c) If f n → f a.e. and f = g a.e., then f n → g a.e.
(d) If f n → f and f = g a.e., then f n → g .
Proof. (a) Let A1 = {x : f n ( x) →
/ f ( x)} and A2 = { x : f n (x ) →
/ g(x )}. Because f n → f
a.e. and f n → g a.e., we have (A1 ) = 0 = (A2 ). Thus, (A1 ∪ A2 ) = 0 also. Then for
all x ∈ X − (A1 ∪ A2 ), we have f n (x ) → f (x ) and f n (x ) → g( x) . By uniqueness of
limits, we have f ( x) = g( x) for all x ∈ X − (A1 ∪ A2 ). Thus, f = g a.e.
Dr. Neal, WKU
(b) Let A = { x ∈ X : f ( x) − g( x) > 0}. We must show that
If f (x ) − f n ( x) <
/ 2 and f n ( x) − g( x) <
f (x ) − g(x ) ≤
Thus, { x : f (x ) − f n (x ) <
(A) = 0.
/ 2 , then we have
f ( x) − f n ( x) + f n (x ) − g( x) < .
/ 2} ∩ { x : f n (x ) − g(x ) <
/ 2} ⊆ { x : f ( x) − g( x) < }.
By taking complements we have
{ x : f (x ) − g( x) ≥ } ⊆ { x : f ( x) − f n (x ) ≥
/ 2} ∪ { x : f n (x ) − g( x) ≥
/ 2}.
Let ′ > 0 be given. Because f n → f and f n → g , there exists N ≥ 1 such that
( f − f N ≥ / 2 ) < ′ / 2 and
( f N − g ≥ / 2 ) < ′ / 2.
Hence, ( f − g ≥ ) ≤ ( f − f N ≥ / 2 ) + ( f N − g ≥
Thus, ({ f − g ≥ }) = 0 for every > 0. But then
A=
/ 2 ) < ′ for every
′ > 0.
U { f − g ≥ },
∈Q +
which is a denumerable union of sets of measure 0. Thus,
(c) Let A1 = {x : f n ( x) →
/ f ( x)} and A2 = { x : f (x ) ≠ g(x )}.
(A) = 0.
Then
(A1 ∪ A2 ) = 0 and
c
for x ∈ ( A1 ∪ A2 ) , we have f n (x ) → f (x )= g( x) . Thus, f n → g a.e.
(d) Let An = {x : f n (x ) − g(x ) ≥ } . Then An = ( An ∩ { f = g}) ∪ ( An ∩ { f ≠ g}) , so that
(An ) = ( An ∩ { f = g}) + (An ∩ { f ≠ g})
≤ ( {x : f n (x ) − g(x ) ≥ } ∩ { f = g}) + ({ f ≠ g})
= ( {x : f n (x ) − f ( x) ≥ } ∩ { f = g}) + 0
≤ ( {x : f n (x ) − f (x ) ≥ }) → 0 as n → ∞ .
Thus, f n → g .
QED
Of the four modes of convergence that we have defined, uniform convergence is
the strongest. If f n ⇒ f , then f n → f which in turn implies f n → f a.e. The next
result shows that f n ⇒ f also implies f n → f .
Dr. Neal, WKU
Theorem 5.7. Let ( X, F , ) be a measure space. If { f n } are measurable functions that
converge uniformly to f on X , then { f n } converges in measure to f .
Proof. Let > 0 be given. Because f n ⇒
f n ( x) − f ( x) < for all x ∈ X when n ≥ N .
makes ( f n − f ≥ ) = 0 . Hence, lim (
n →∞
f , there exists an integer N ≥ 1 such that
Thus, for n ≥ N , { f n − f ≥ } = ∅ which
f n − f ≥ ) = 0 for all > 0 and f n → f .
QED
Convergence in measure does not imply uniform convergence nor does it imply
convergence a.e. However, we can say the following:
Theorem 5.8. If f n → f , then there exists a subsequence such that f nk → f a.e.
Proof. Because f n → f , there exists an increasing sequence of integers {nk } such that
(
)
{
}
f n − f ≥ 2 −k < 2− k for all n ≥ nk . Now let Ek = x ∈ X : f nk (x ) − f ( x) ≥ 2 −k , and
∞
∞
∞
∞
∞
∞
{
i=1 k =i
f nk − f < 2 −k
let A = I U Ek . Suppose x ∈ A c = U I Ekc = U I
i =1 k = i
i=1 k =i
be given.
∞
{
k=i
Because x ∈ A c , there is some i ≥ 1 such that x ∈ I
>0
}
f nk − f < 2−k . Then for
K ≥ i large enough, we have 2 −K < ; so, for k ≥ K we have
Thus, f nk → f for all x ∈ A c . Moreover, we assert that
} and let
f nk (x ) − f (x ) < 2 −k < .
(A) = 0.
For each i ≥ 1, we have
(A) =
 ∞ ∞

I UE ≤
k

 i =1 k =i 
 ∞

i
∞
∞
k
 U E  ≤ ∑ ( E ) ≤ ∑  1  = (1 / 2) = (1/ 2) i −1 .
k
k

 
1 − 1/ 2
k =i  k =i
k=i 2
Because 0 ≤ ( A) ≤ (1/ 2) i −1 for all i ≥ 1, we must have
(A) = 0. Thus, f nk → f a.e.
QED
Finally, convergence almost everywhere does not imply uniform convergence nor
pointwise convergence. However, it will imply convergence in measure provided we
have a finite measure space.
Dr. Neal, WKU
Theorem 5.9 (Egorov). Let ( X, F , ) be a finite measure space. If f n → f a.e., then
fn → f .
Proof.
Let
> 0 be given.
A = { x ∈ X : f n ( x) →
/ f ( x)} .
c
(A ) =
( X) − ( A) =
∞
We must show that
Because f n → f
( fn − f ≥ ) = 0.
a.e., we have
Let
(A) = 0 and thus
(X ).
For all n ≥ 1, I { f i − f < } ⊆
i =n
lim
n →∞
∞
∞
i = n+1
i =n
I { f i − f < } ; thus the sets I { f i − f < }
form a nested increasing sequence of events as n increases. Therefore,
 ∞ ∞

 ∞

 U I { f − f < } = lim  I { f − f < } .
i
i

 n→∞ 

 n =1 i = n

i= n

∞
However, if x ∈ A c , then f n (x ) → f (x ); thus, x ∈ U
∞
I { f i − f < }.
Hence,
n=1 i = n
∞
∞
Ac ⊆ U I { f i − f < } and therefore
n =1 i = n
 ∞

 I { f − f < } =
i

n→ ∞  i = n

 ∞ ∞

 U I { f − f < } ≥
i


 n =1 i = n

(X ) ≥ lim
Because
(A c ) =
( X) .
 ∞
c 


(X ) < ∞, we must have lim   I { f i − f < }  = 0 . Therefore,

n →∞   i = n
 
 ∞


lim ( { f n − f ≥ }) ≤ lim  U { f i − f ≥ } = 0 .
n →∞
n→ ∞  i = n

QED
The following definition gives another version of the concept of ”Cauchy.”
Definition 5.2. A sequence of measurable functions { f n } is called Cauchy in measure (or
fundamental in measure) provided for every > 0 we have lim
( fn − fm ≥ ) = 0.
n, m →∞
(That is, for every
> 0 and every
′ > 0 there exists an integer N ≥ 1 such that
( f n − f m ≥ ) < ′ whenever n,m ≥ N .
Theorem 5.10. If f n → f , then { f n } is Cauchy in measure.
Dr. Neal, WKU
Proof.
Let
such that
> 0 and ′ > 0 be given. Because f n → f , there exists an integer N ≥ 1
( fn − f ≥
/ 2) < ′ / 2 for all n ≥ N . Then for all n, m ≥ N we have
{ fn − f <
/ 2} ∩ { f − f m <
/ 2} ⊆ { f n − f m < };
thus, by taking complements { f n − f m ≥ } ⊆ { f n − f ≥
/ 2} ∪ { f − f m ≥
/ 2}.
So for n, m ≥ N we have
( fn − fm ≥ ) ≤
( fn − f ≥
/ 2) + ( f − f m ≥
/ 2) < ′ / 2 + ′ / 2 = ′ .
Hence, { f n } is Cauchy in measure.
QED
Convergence in Distribution
Let (Ω, F , P ) be a probability space and let {X n}∞
n =1 be a sequence of random variables
defined on Ω . We say that Xn converges in distribution to the random variable X
provided lim P( Xn ≤ t ) = P( X ≤ t) for all t ∈ℜ .
n →∞
It follows that
lim P(s ≤ Xn ≤ t) = P(s ≤ X ≤ t) for all s, t ∈ℜ .
n →∞
Although this
concept can be generalized to any measure space, it is used mostly in probability and
statistics to approximate probabilities.
Example 5.3. (a) Let X be any random variable and let Xn be the truncation of X
defined by
 X if X ≤ n
Xn = X 1{X≤n} = 
,
 n if X > n
for n ≥ 1. Then Xn converges in distribution to X . To see this, let t ∈ℜ . Then let t < n .
If X ≤ t , then X ≤ n so Xn = X and Xn ≤ t also. Moreover, if Xn ≤ t then Xn < n so we
cannot have X > n . Thus, X = X n and X ≤ t also. That is, X ≤ t iff Xn ≤ t for t < n . So
for n > t we have P(X n ≤ t) = P( X ≤ t ).
(b) As a specific example of truncation, let X ~ geo( p) be a geometric random variable
that counts the number of attempts needed for the first win, where p is the probability
of a win on any independent attempt and q = 1 − p is the probability of a loss.
If you do not win within n attempts, then you quit so that you make a maximum of
n attempts. Then Xn = X 1{X≤n} gives the number of attempts needed to win if it takes
no more than n tries, or gives n when it takes more than n tries to win.
The probability distribution functions are given by
P(X = k ) = q
k−1
p for k ≥ 1
and
q k−1p
P(X n = k ) = 
 q n−1
if 1 ≤ k ≤ n − 1
if k = n
.
Dr. Neal, WKU
(If you have n − 1 losses in a row, with probability q n−1 , then it automatically takes at
least n tries; thus, Xn = n .)
The cumulative distribution functions are given by
 0
P(X ≤ k) = 
1 − q k
if k ≤ 0
if k ≥ 1
and
 0
if k ≤ 0

P(X n ≤ k ) = 1 − q k if 1 ≤ k ≤ n − 1 .

 1
if k ≥ n
For t ∈ℜ , P(X ≤ t ) = P(X ≤  t  ) and P(X n ≤ t) = P( Xn ≤  t  ) . So for t ∈ℜ , by choosing
n > t , we then have  t  ≤ n − 1; thus, P(X n ≤ t) = 1 − qt  = P( X ≤ t) for t ≥ 1 and
P(X n ≤ t) = 0 = P( X ≤ t) for t < 1.
(c) Let {X n} and X be random variables such that 0 ≤ X n ≤ Xn+1 ≤ X for all n ≥ 1 and
lim Xn ( ) = X( ) for all ∈Ω . We assert that {X n} converges in distribution X .
n →∞
If X ( ) ≤ t , then Xn+1 ( ) ≤ t which in turn implies that Xn ( ) ≤ t . Thus, we have
{X ≤ t} ⊆ {Xn+1 ≤ t} ⊆ {Xn ≤ t} for all n ≥ 1, which makes {X n ≤ t} a nested, decreasing
sequence of sets. Because a probability space is a finite measure space, we then have
 ∞


that P  I {Xn ≤ t} = lim P( Xn ≤ t ).
 n =1
 n→ ∞
But if
∞
∈ I {Xn ≤ t}, then Xn ( ) ≤ t for all n ; thus, X ( ) = lim X n ( ) ≤ t . Hence,
n =1
n→∞
 ∞

Thus, P(X ≤ t ) = P  I {Xn ≤ t} = lim P( Xn ≤ t) for all t ∈ℜ ;
n =1
 n =1
 n →∞
hence, {X n} converges in distribution X .
∞
I {Xn ≤ t} = {X ≤ t} .
The Truncation of a Function into Simple Functions
This next construction gives a further generalization of truncation and is crucial to the
development of integration theory.
Initially, let f : X → [0, ∞) be a non-negative real-valued function. We will construct
a sequence of simple functions { f n } such that for all x ∈ X
0 ≤ f1 ( x) ≤ f 2 (x ) ≤ . .. ≤ f n ( x) ≤ . . . ≤ f ( x)
and
lim f n ( x) = f (x ).
n →∞
That is, the sequence { f n } increases pointwise to f .
For all n ≥ 1, we shall let f n (x ) = n whenever f ( x) ≥ n . Next, suppose 0 ≤ f (x ) < n.
Then k ≤ f ( x) < k + 1 for some integer k with 0 ≤ k < n − 1. Now we divide this interval
i−1
i
[k , k + 1) into 10 n intervals of equal length. Then we have k + n ≤ f ( x) < k + n for
10
10
n
some integer i with 1 ≤ i ≤ 10 .
Dr. Neal, WKU
To define the required simple functions, we let En = {x : f (x ) ≥ n} for n ≥ 1. Next,

i−1
i 
we let Ek,i, n =  x : k + n ≤ f (x) < k + n  for k = 0, .. ., n − 1 and i = 1,... , 10n , and let

10
10 
fn =
n−1 10 n 
i − 1
∑ ∑  k + 10 n  1Ek,i,n + n1En .
k=0 i=1
For example, suppose 4 ≤ f (x ) < 5.
Then f 1 (x ) = 1, f 2 (x ) = 2 , f 3 (x ) = 3, and
f 4 (x ) = 4 . For f 5 ( x) , we divide the interval [4, 5) into 10 5 = 100, 000 subintervals of
the form [4 + (i − 1) / 10 5 , 4 + i / 105 ) with the endpoints being
4.00000 , 4.00001 , 4.00002 , . . . , 4.00009 , 4.00010 , . . . , 4.00199 , . . . , 4.99999 , 5
Then f ( x) is in exactly one of these subintervals and we round f 5 ( x) down to the left
endpoint 4 + (i − 1) / 10 5 . So f 5 ( x) = 4. a1 a2 a3 a4 a5 which is the truncation of the
decimal expansion of f ( x) to 5 decimal places.
Then for n ≥ 5 , f n (x ) = 4. a1 a2 a3 a4 . . . .an which is the truncation of the decimal
expansion of f ( x) to n decimal places. Then 0 ≤ f n (x ) ≤ f n+1( x) ≤ f ( x) for all x ∈ X
and lim f n ( x) = f (x ).
n →∞
Note: (a) If f is measurable with respect to a measure space ( X, F , ) , then each set
En and Ek,i, n are in F ; thus, each indicator function in f n will be measurable and thus
each f n will be measurable.
(b) For an arbitrary f , we can apply the construction to each of f + and f − obtaining
sequences of simple functions {hn} and {gn } for each respectively.
Then let
f n = hn 1{f ≥ 0} − gn 1{ f < 0} . Then { f n } are still simple functions such that for all x ∈ X
lim f n ( x) = f (x ). Moreover, when f ( x) ≥ 0 then 0 ≤ f n (x ) ≤ f n+1( x) ≤ f ( x) , and
n →∞
when f ( x) < 0 , then 0 ≥ f n (x ) ≥ f n+1( x) ≥ f ( x) .
(c) If f ≥ 0 and we are working on a finite measure space ( X, F , ) , in particular a
probability space, then as in Example 5.3 (c), we would have that
lim ( f n ≤ t ) = ( f ≤ t) for all t ∈ℜ .
n →∞
The preceding construction gives us the following result:
Theorem 5.11. Let ( X, F , ) be a measure space and let f : X → ℜ be a real-valued
function defined on X . There exists a sequence of simple functions { f n } that converges
pointwise to f on X . If f is measurable, then the functions { f n } may be chosen to be
measurable. If f ≥ 0 , then { f n } may be chosen so that 0 ≤ f1 ≤ f 2 ≤ . . . ≤ f n ≤ . . . ≤ f .
Dr. Neal, WKU
Additional Exercises
Exercise 5.5. Let ( X, F , ) be a finite measure space. Assume f n → f and gn → g .
(a) Let a , b be non-zero constants. Prove that a f n + bg n → a f + b g.
(b) Prove that f n →
f .
(c) Prove that f n2 → f 2 .
(d) Prove that f n gn → f g . (Hint: Apply the identity that expresses a product in terms
of sums and squares.)
Exercise 5.6. Let ( X, F , ) be a measure space. For measurable functions f , g , say
f ~ g provided f = g a.e. Prove that ~ is an equivalence relation.