Problem set 2, Real Analysis I, Spring, 2015.
(18) Prove the following assertion: Every measurable function is the
limit a.e. of a sequence of continuous functions.
[Hint: Let ψ be a step function, and let be positive. Show
there is a continuous function which is equal to ψ outside a set
of measure .]
P
Solution: First of all, we prove the hint: Let ψ = N
j=1 aj χRj
be a step function, where the Rj are closed rectangles. Each
rectangle in Rd is a product of intervals of the form [a, b] ⊂ R.
For χ[a,b] , consider




1
0
g [a,b] (x) =
1
(x − a + )


 1 − 1 (x − b)
if
if
if
if
a ≤ x ≤ b,
x ≥ b + or x ≤ a − ,
a − < x < a,
b < x < b + .
which is continuous and equal to χ[a,b] outside a set of measure
2.
For rectangle R = [a1 , b1 ] × · · · × [ad , bd ] in Rd , then χR is
equal to g R (x) = Πdi=1 g [ai ,bi ] (xi ) on a outside a set of measure
C(), where C() ≤ 2[maxi (bi − ai ) + 2]d−1 , which goes to 0
as → 0.
Since ψ is a finite linear combination
of N characteristic funcP
Rj
tions of rectangles, we see ψ = N
a
except on a set of
j=1 j g
measure m which goes to zero as → 0. This proves the hint.
Now let f be a measurable function. Then there is a sequence
ψk of step functions which converges to f for all x ∈ Rd − Y ,
where m(Y ) = 0. Now let gk be a continuous function which is
equal to ψk outside a set Ek of measure at most 2−k . Now if
x∈
∞
∞ \
[
(Ekc − Y ),
n=1 k=n
then there is an n so that if k ≥ n, ψk (x) = gk (x). So for such
x, f (x) = limk→∞ ψk (x) = limk→∞ gk (x). So define
E=
"
∞ \
∞
[
#c
(Ekc − Y )
n=1 k=n
=Y ∪
∞ [
∞
\
n=1 k=n
1
Ek
!
.
2
So compute the measure of E as
∞ [
∞
\
m(E) ≤ m(Y ) + m
Ek
!
n=1 k=n
≤ m
∞
[
Ek
!
for all n
k=n
≤
≤
∞
X
k=n
∞
X
m(Ek )
2−k = 21−n .
k=n
Letting n → ∞, we see m(E) = 0, and for x ∈
/ E, gn (x) → f (x).
This completes the proof.
(37) Suppose Γ is a curve y = f (x) in R2 , where f is continuous.
Show that m(Γ) = 0.
[Hint: Cover Γ by rectangles, using the uniform continuity of
f .]
Solution:
Let Γn = {(x, y) : y = f (x), x ∈ [−n, n]. Since
S∞
Γ = n=1 Γn , it suffices to show m(Γn ) = 0 for each n.
Fix n. Since [−n, n] is compact, f is uniformly continuous
on [−n, n]. Therefore, for all > 0, there is a δ > 0 so that
if |x − y| ≤ δ for x, y ∈ [−n, n], then |f (x) − f (y)| ≤ . Now
if m > 1δ , then consider the 2mn intervals of the form Ik =
k
[−n + m
, −n + k+1
]. So the uniform continuity implies that if
m
k
x ∈ Ik , then |f (x) − f (−n + m
)| ≤ . So let
Rk = Ik × [f (−n +
k
)
m
− , f (−n +
k
)
m
+ ]
S
be a rectangle in R2 . Our construction implies Γn ⊂ 2mn
k=1 Rk .
Now for all > 0,
! 2mn
2mn
2mn
[
X
X
1
m(Γn ) ≤ m
Rk =
m(Rk ) =
· 2 = 4n.
m
k=1
k=1
k=1
Let → 0. Therefore, m(Γn ) = 0 for all n and so m(Γ) = 0.
(3) Let {an } be a sequence of extended real numbers. L ∈ [−∞, ∞]
is said to be a limit point of the sequence if there is a subsequence ank so that limk→∞ ank = L. Show that inf k supn≥k an is
equal to the supremum of the set of limit points of {an }. This
quantity is defined as lim sup an . Show that lim sup an is a limit
point of {an }.
3
Solution: Let K be the supremum of the set of limit points.
Let P = K − if K ∈ R, and P = N if K = ∞, and P = −∞ if
K = −∞. So there is a subsequence anj so that limj→∞ anj ≥
P . This implies that for all k,
sup an ≥ sup anj ≥ lim anj ≥ P.
n≥k
nj ≥k
j→∞
Therefore, inf k supn≥k an ≥ P . Now let P → K to show
inf k supn≥k an ≥ K.
Conversely, let M = inf k supn≥k an . Then for all k, supn≥k an ≥
M . Now we construct a subsequence recursively. Let Qj = j if
M = ∞, Qj = M − 1j if M ∈ R, and Qj = −∞ if M = −∞.
Then Qj % M in each case. So define n1 so that an1 ≥ Q1 ,
and for j > 1, define anj so that nj ≥ nj−1 and anj ≥ Qj . This
is always possible since supn≥k an ≥ M in for all k. Now anj
may not be convergent, but its values lie in the compact topological space [Q1 , ∞] (which is homeomorphic to the compact
metric space [0, 1]). This implies there is a convergent subsequence anj` , and the conditions anj ≥ Qj with Qj → M shows
that K ≥ lim`→∞ anj` ≥ M . But the previous paragraph shows
M ≥ K. Thus K = M , and moreover K is a limit point of
{an }.
(4) The Heaviside function H(x) is defined to be the characteristic
function χ[0,∞) .
(a) Show that there is no continuous function f that is equal
to H almost everywhere.
Solution: If on the contrary, there is a continuous f so
that f = H a.e., consider the open interval I = (f (0) −
1
, f (0) + 14 ). Since f is continuous, f −1 (I) is an open set
4
containing 0. In particular, the open nonempty sets O1 =
f −1 (I) ∩ (0, ∞) and O2 = f −1 (I) ∩ (−∞, 0) each have
positive measure. Therefore, there are points xi ∈ Oi for
i = 1, 2 so that f (xi ) = H(xi ). Now H(x1 ) = 0 and
H(x2 ) = 1. By our construction of I,
|H(x1 ) − H(x2 )| = |f (x1 ) − f (x2 )| < 21 .
This is a contradiction.
(b) Show that there is no sequence fk of functions in C 0 (R) so
that fk → H with respect to the L∞ (R) norm.
Solution: Recall that there is a natural norm-preserving
injection Φ : C 0 (R) → L∞ (R). So we may view C 0 (R) as
a subset L∞ (R) with the induced norm from L∞ (R). We
4
know C 0 (R) and L∞ (R) are both Banach spaces, in that
the metrics induced from the norms are complete. The
problem then follows from part (a) and the following standard fact:
Let X ⊂ Y for Y a metric space, and consider X as a
metric space with the induced metric from Y . Assume X
is a complete metric space. Then X is a closed subspace
of Y .
So, keeping this in mind we provide a proof by contradiction. If fk → H in L∞ (R), then fk is a Cauchy sequence in L∞ (R). By the norm-preserving injection of
C 0 (R) → L∞ (R), we see fk is a Cauchy sequence in C 0 (R)
as well. Therefore, since C 0 (R) is complete, there is a continuous function g so that fk → g in C 0 (R) and so fk → g
in L∞ (R) as well. But this implies g = H in L∞ (R), which
is equivalent to g = H almost everywhere. But g is continuous, and part (a) provides a contradiction.
(5) A function f : Rd → R is said to vanish at infinity if lim|x|→∞ f (x) =
0. Show that the closure in L∞ (Rd ) of the set of continuous Rvalued functions with compact support on Rd is equal to the
set of continuous functions which vanish at infinity on Rd .
Solution: Let f ∈ L∞ (Rd ) be equal to the limit f = limk→∞ gk
in L∞ , where each gk is continuous with compact support. As
in (4b) above, we must have f ∈ C 0 (Rd ) and
kgk − f kC 0 = sup |gk (x) − f (x)| → 0.
So we need to show that lim|x|→∞ f (x) = 0. In other words, we
want to show that for all > 0, there is an N so that if |x| ≥ N ,
then |f (x)| ≤ .
We show lim|x|→∞ f (x) = 0 by contradiction. If this were
not true, then there would be an > 0 so that for all N , there
is an xN with |xN | ≥ N so that |f (xN )| > . Since each gk
has compact support, there is an Nk so that if |x| ≥ Nk then
g(x) = 0. Therefore,
kgk − f kC 0 = sup |gk (x) − f (x)| ≥ |gk (xNk ) − f (xNk )| > .
This is a contradiction, and thus f (x) → 0 as |x| → ∞.
Now by converse, we must show that for f ∈ C 0 (Rd ) with
lim|x|→∞ f (x) = 0, that there is a sequence gk ∈ C 0 (Rd ) each
with compact support so that
kgk − f kL∞ = kgk − f kC 0 → 0.
5
So let φk be a continuous function with compact support which
has range [0, 1], is identically 1 on Bk (0) and is identically 0 on
Bk+1 (0)c . For example, we can choose

1
if |x| ≤ k

k + 1 − |x| if k < |x| < k + 1
φk (x) =

0
if |x| ≥ k + 1
Then each φk is continuous and φk % 1 as k → ∞.
Let gk = f · φk . Note
|f (x) − gk (x)| = |(1 − φk (x))f (x)| = |1 − φk (x)| · |f (x)|.
Now let > 0. Then there is an N so that if |x| ≥ N , then
|f (x)| ≤ . Since 0 ≤ 1−φk (x) ≤ 1, this implies |f (x)−gk (x)| ≤
for |x| ≥ N . But then for k ≥ N , gk (x) = f (x) for |x| ≤ k.
Therefore, if k ≥ N , then sup |gk − f | ≤ . This implies gk → f
in L∞ (Rd ).