MATH 660A: SOLUTION TO TWO HOMEWORK
PROBLEMS
WINFRIED JUST, OHIO UNIVERSITY
Theorem 1. Suppose {fn } be a sequence of continuous functions with fn :
En → R for all n and let f : R → R+ be such that for every x ∈ R there
exists Nx such that ∀n > Nx x ∈ En and limn→∞ fn (x) = f (x). Then f is
measurable.
Proof: Since R+ is homeomorphic with [0, 1], we may without loss of generality assume that f and all fn are uniformly bounded, that is, that the
range of f, fn is contained in some interval [−M, M ].
Lemma 2. For each n there exists a function fn+ with fn+ En = fn such
that fn+ is continuous on all points in En .
Proof: First let us define fn+ on the closure E¯n of En . For each x ∈ Ē
choose a sequence {xm } of elements of En such that limm→∞ xm = x.
Since the range of fn is bounded, by the Bolzano-Weierstrass Theorem
we may choose {xm } in such a way that limm→∞ fn (xm ) exists. Define
f + (x) := limm→∞ fn (xm ). We need to show that the function fn+ E¯n is
still continuous on En . Pick x ∈ En and let ε > 0. Then there exists δ > 0
such that |f (y)−f (x)| < 2ε for all y ∈ En with |y−x| < δ. Now if y ∈ E¯n satisfies |y −x| < δ and y = limm→∞ xm with xm ∈ En , then |xm −x| < δ for all
but finitely many m, and it follows that |f (y)−f (x)| ≤ 2ε < ε. Thus fn+ E¯n
is continuous on En . Now consider a family of pairwise
disjoint open interS
c
vals (ai , bi ) with endpoints in E¯n such that E¯n = i∈N (ai , bi ). If ai = −∞
we define fn+ (ai , bi ) = fn+ (bi ); if bi = ∞ we define fn+ (ai , bi ) = fn+ (ai );
otherwise we define fn+ [ai , bi ] as a linear function with values at ai , bi
as already specified. It is straightforward to see that this function remains
continuous at all points in En . Now let us derive the theorem from the lemma. Let {fn } be as in the
lemma, and for each n let Cn be the set of all points where fn+ is continuous.
Then En ⊆ Cn for all n and Cn is a Gδ -set of reals by a previous
homework
S
exercise. Now let us define for every k ∈ N the set Dk = n≥k Dn . Then
each set Dk is an intersection of countably many Borel sets, hence Dk is
Borel and thus measurable. Moreover, ∀k ∈ N Dk ⊆ Dk+1 . Thus for each
k ∈ N the sequence {fn+ Dk : n ≥ k} is a sequence of measurable functions
that converges pointwise to f Dk . By Proposition 9 of the textbook, for
all k ∈ N the function f Dk is measurable.
1
2
WINFRIED JUST, OHIO UNIVERSITY
Moreover, byT
the assumptions
T of the theorem for each x ∈ R there exists
k ∈ N with x ∈ n≥k En and n≥k En ⊆ Dk . Therefore we are assured that
S
k∈N Dk = R.
Now let B ⊆ R be Borel. Then (f Dk )−1S
[B] is measurable for all k, and
thus f −1 [B] is also measurable as f −1 [B] = k∈N (f Dk )−1 [B]. Theorem 3. Let {fn } be a sequence of measurable functions on E. Then
{fn } is uniformly integrable and tight over E iff for each ε > 0, there is a
measurable subset E0 of E that has finite measure and a δ > 0 such that for
each measurable subset A of E and index n,
Z
(1)
m(A ∩ E0 ) < δ ⇒
|fn | < ε.
A
Proof: For the “only if”-direction (⇒) assume that {fn } is uniformly integrable and tight over E. Fix ε > 0. By
R the assumption on tightness,
there exists E0 ⊆ E of finite measure with E∼E0 |fn | < 2ε for each index n.
By uniform integrability, there exists
R δ > 0 such that for all A ⊆ E with
m(A) < δ and all n ∈ N we have A |fn | < 2ε . Then for each n ∈ N and
measurable A ⊆ E with m(A ∩ E0 ) < δ we have
Z
Z
|fn | =
A
Z
|fn | +
A∩E0
Z
|fn | ≤
A∼E0
Z
|fn | +
A
|fn | <
E∼E0
ε ε
+ = ε.
2 2
For the “if”-direction (⇐) assume that δ > 0 and a subset E0 of E that
has finite measure are such that for each measurable subset A of E and
index n the implication (1) holds. RFor A = E ∼ E0 we have m(A ∩ E0 ) =
m(∅) = 0 < δ, and (1) implies that E∼E0 |fn | < ε for all n ∈ N, which is the
definition of tightness. On the other hand, if m(A) < δ, then m(A∩E0 ) < δ,
and (1) gives uniform integrability.