A note on some approximation theorems in measure theory

S. Kesavan and M. T. Nair

Department of Mathematics,
Indian Institute of Technology, Madras,
Chennai - 600 036
email: kesh@iitm.ac.in and mtnair@iitm.ac.in

Abstract

Usual proofs of the density of continuous functions with compact support in

the Lebesgue spaces Lp (Ω), where 1 ≤ p < ∞, and Ω is an open subset of
RN , appeal to Lusin’s theorem. It is shown here that this density theorem
can be proven directly and that Lusin’s theorem can be deduced from it.

1
1 Introduction
One of the important approximation theorems in the theory of measure and
integration is the density of continuous functions with compact support in
the Lebesgue spaces Lp (Ω), where Ω is an open subset of RN and 1 ≤ p < ∞.
This result, together with the technique of convolution with mollifiers, shows
that infinitely differentiable functions with compact support in Ω are also
dense in Lp (Ω), 1 ≤ p < ∞. Further, separability properties of these spaces
can also be deduced from this result. These are just some of the applications
of this theorem.
Usual proofs of this approximation theorem appeal to Lusin’s theorem
which states that a measurable function on a set of finite Lebesgue measure
agrees with a continuous function with compact support except possibly on
a set whose measure is less than an arbitrarily small preassigned positive
number.
In this note we will show that we can actually prove the density theo-
rem by starting from simple results based on approximation of measurable
functions by step functions and that we can deduce Lusin’s theorem as a
consequence.
Throughout the sequel, we will assume that RN is equipped with the
Lebesgue measure (denoted µ). Let 1 ≤ p < ∞. If Ω ⊂ RN is an open subset,
then Lp (Ω) will denote the space (of equivalence classes, up to equality almost
everywhere) of real (or complex) valued functions f defined on Ω with the
property Z
|f |p dx < ∞.
Ω
The norm on this space is denoted k · kp and is defined by
Z  p1
kf kp = |f |p dx .
Ω

The essential supremum of an essentially bounded function f is denoted by

kf k∞ . The space of continuous functions with compact support in Ω will
be denoted by Cc (Ω) and will be equipped with the usual sup-norm, also
denoted by k · k∞ .
By a simple function, we will mean a function of the form
k
X
f= αi χEi , (1.1)
i=1

2
where, for 1 ≤ i ≤ k, the αi are real (or complex) numbers and χEi are
the characteristic (also called indicator) functions of the measurable subsets
Ei ⊂ Ω.
By a box in RN , we will mean a set of the form
ΠN
j=1 Ij

where the Ij are all finite intervals in R. A closed box is the corresponding
product of closed intervals and an open box is the product of open intervals.
A half-open box is a box of the form
ΠN
j=1 [aj , bj ).

Lemma 1.1 Every open set in RN can be written as a countable disjoint

union of half-open boxes.
Proof: For a positive integer n, let Pn denote the collection of all points in
RN whose coordinates are all integral multiples of 2−n . Let Bn denote the
collection of all half-open boxes with each edge of length 2−n and with vertices
at the points of Pn . The following conclusions are obvious by inspection:
(i) For a fixed n, each point x ∈ RN belongs to exactly one box in Bn .
(ii) If Q ∈ Bm , Q0 ∈ Bn where n > m, then either Q0 ⊂ Q or Q ∩ Q0 = ∅.
Let Ω ⊂ RN be an open set. Let x ∈ Ω. Then x lies in an open ball
contained in Ω and so we can find a Q ∈ Bn , for some sufficiently large n,
containing x and which is contained inside this ball. In other words, Ω is the
union of all boxes contained inside it and belonging to some Bn . This col-
lection of boxes is clearly countable but may not be disjoint. Now choose all
those boxes in this collection which are in B1 and discard boxes of Bk , k ≥ 2,
which lie inside these selected boxes. From the remaining collection of boxes,
select those in B2 and discard those boxes in Bk , k ≥ 3, which lie inside these
selected boxes, and so on. The lemma now follows from observations (i) and
(ii) above. 

By a step function, we mean a simple function as in (1.1), where the

sets Ei , 1 ≤ i ≤ k, are all boxes.

2 Some approximation results

In this section, we prove some preliminary lemmas from which we will deduce
the approximation results mentioned earlier.

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Lemma 2.1 Let Ω ⊂ RN be an open set and let E ⊂ Ω be a set of finite
measure. Let ε > 0 be given. Then, there exists a set F , which is a finite
disjoint union of boxes, such that

µ(E∆F ) < ε,

where E∆F = (E\F ) ∪ (F \E).

Proof: Since µ(E) < ∞, there exists an open set G ⊂ RN such that E ⊂ G
and µ(G\E) < 2ε . Set G0 = Ω ∩ G. Then G0 is also open, E ⊂ G0 and
µ(G0 \E) < 2ε . Now G0 can be written as a countable Pdisjoint union of boxes
∞
{Ij }∞
j=1 and we consequently have Ij ⊂ Ω. Since j=1 µ(I j < ∞, let us
)
choose k such that ∞
X ε
µ(Ij ) < .
j=k+1
2

Set F = ∪kj=1 Ij , which is a finite disjoint union of boxes and we also have
F ⊂ G0 . Now
ε
µ(F \E) ≤ µ(G0 \E) <
2
and ∞
X ε
µ(E\F ) ≤ µ(G0 \F ) ≤ µ(Ij ) < .
j=k+1
2

This completes the proof. 

Lemma 2.2 Let E ⊂ RN be a set of finite measure. Let ε > 0 be given.

Then there exists a compact set K ⊂ E such that

µ(E\K) < ε.

Proof: Step 1: Since E has finite measure, there exists an open set V such
that E ⊂ V and µ(V \E) < ε.

Let B(0; r) denote the open ball centred at the origin and of radius r.
The corresponding closed ball will be denoted B(0; r). For a positive integer
n, set Vn = B(0; n) ∩ V . Then Vn ↑ V and so there exists a positive integer
m such that µ(V \Vm ) < ε.Then

E∆Vm = (E\Vm ) ∪ (Vm \E) ⊂ (V \Vm ) ∪ (V \E)

4
and so µ(E∆Vm ) < 2ε.

Step 2: Now Vm is a bounded open set and so there exists R > 0 such that
Vm ⊂ B(0; R). Then, we can find ρ > 0 such that µ(B(0; R)\B(0; ρ)) < ε.
Further, there exists a closed set F ⊂ Vm such that µ(Vm \F ) < ε. Conse-
quently, B(0; ρ) ∩ F is compact and

Vm \(B(0; ρ) ∩ F ) = (Vm \B(0; ρ)) ∪ (Vm \F )

⊂ (B(0; R)\B(0; ρ)) ∪ (Vm \F )

and so µ(Vm \(B(0; ρ) ∩ F )) < 2ε.

Step 3: If E ⊂ RN is of finite measure, we have seen in Step 1 above that there

exists a bounded open set W such that µ(E∆W ) < 3ε . We have also seen
in Step 2, that there exists a compact set K e ⊂ W such that µ(W \K) e < ε.
3
ε
Finally, there exists a closed set F ⊂ E such that µ(E\F ) < 3 . Then
e e
K = Fe ∩ K e is compact, K ⊂ E and

E\K = (E\W ) ∪ ((E ∩ W )\Fe) ∪ ((W ∩ Fe)\K)

e

from which it follows that µ(E\K) < ε. This completes the proof. 

Lemma 2.3 Let I ⊂ RN be a box and let ε > 0 be given. Then there exists
ϕ ∈ Cc (RN ) such that 0 ≤ ϕ(x) ≤ 1 for all x and

µ({x ∈ RN | ϕ(x) 6= χI (x)}) < ε.

Further, the support of ϕ will be contained in I.

Proof: Choose a closed box J1 and an open box J2 such that J1 ⊂ J2 ⊂ J2 ⊂

I and such that µ(I\J1 ) < ε. By Urysohn’s lemma, there exists a continuous
function ϕ such that 0 ≤ ϕ(x) ≤ 1 for all x and such that ϕ(x) = 1 for all
x ∈ J1 and ϕ(x) = 0 for all x ∈ RN \J2 . Then the support of ϕ is contained
in J2 ⊂ I, which is compact and so ϕ ∈ Cc (RN ). Now,

{x ∈ RN | ϕ(x) 6= χI (x)} ⊂ I\J1

and the result now follows immediately. 

5
Remark 2.1 Urysohn’s lemma is a result valid in the fairly general setting of
a topological space which is normal (cf. Simmons [1]). It is much simpler in
the context of a metric space and even simpler in RN . For example if we are
in the real line and have a closed interval [a, b] contained in an open interval
(c, d), then we can easily construct a continuous function, which is identically
equal to unity on [a, b] and which vanishes outside (c, d), as follows:

 0, if x ≤ c or x ≥ d,
 x−c , if c ≤ x ≤ a,

ϕ(x) = a−c

 1, if a ≤ x ≤ b,
 d−x
d−b
, ifb ≤ x ≤ d.

A similar construction can be done in the case of boxes in RN . Another way

to construct such functions (which is valid in any metric space (X, d)) is to
define, for any set A, the distance of a point x from A by

d(x, A) = inf d(x, y);

y∈A

this function is continuous. Now define

d(x, A)
ϕ(x) = .
d(x, A) + d(x, B)

If A and B are disjoint closed sets, then ϕ is a well-defined continuous func-

tion such that ϕ ≡ 0 on A and ϕ ≡ 1 on B. 

Corollary 2.1 Let Ω ⊂ RN be an open set. Let f be a step function defined

on Ω and let ε > 0 be given. Then, there exists ϕ ∈ Cc (Ω) such that

µ({x ∈ Ω | ϕ(x) 6= f (x)}) < ε (2.1)

and such that

kϕk∞ ≤ kf k∞ . (2.2)

Proof: Let f = kj=1 αi χIj where the Ij are all disjoint boxes contained in
P

Ω. By Lemma 2.2 above, there exists ϕj ∈ Cc (RN ) with support contained

in Ij ⊂ Ω, taking values in [0, 1] and such that
ε
µ({x ∈ RN | ϕj (x) 6= χIj (x)} < .
k

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 Pk
Set ϕ = j=1 αj ϕj . Then

{x ∈ Ω | ϕ(x) 6= f (x)} ⊂ ∪kj=1 {x ∈ Ω | ϕj (x) 6= χIj (x)}

⊂ ∪kj=1 {x ∈ RN | ϕj (x) 6= χIj (x)}

and so
µ({x ∈ Ω | ϕ(x) 6= f (x)}) < ε.
Since the supports of the ϕj are disjoint, it follows that

kϕk∞ ≤ max |αj | = kf k∞ .

1≤j≤k

Observe that the function ϕ has compact support contained in ∪kj=1 Ij ⊂ Ω.

Thus ϕ ∈ Cc (Ω). 

3 Density of Cc(Ω) in Lp(Ω)

Let Ω ⊂ RN be an open set and let 1 ≤ p < ∞. Define

S = {ϕ : Ω → R | ϕ is simple, and µ({x ∈ Ω | ϕ(x) 6= 0}) < ∞}.

Observe that a simple function ϕ belongs to Lp (Ω), for 1 ≤ p < ∞, if, and
only if, it belongs to S. Clearly, step functions are in S.

Proposition 3.1 Let S be defined as above. Then S is dense in Lp (Ω), for

1 ≤ p < ∞.

Proof: Let f ∈ Lp (Ω) be a non-negative function. Let {ϕn } be simple func-

tions such that 0 ≤ ϕn ≤ f and such that ϕn ↑ f . Then, clearly ϕn ∈ Lp (Ω)
and so ϕn ∈ S, for each positive integer n. Since |ϕn −f |p ≤ 2p |f |p , and since
|f |p is integrable, it follows from the dominated convergence theorem that
ϕn → f in Lp (Ω). If f is real valued, the result can be deduced by observing
that f = f + − f − where f + = max{f, 0} ≥ 0 and f − = − min{f, 0} ≥ 0. If
f is complex valued, we can apply the preceding arguments to Ref and Imf
to obtain the same conclusion for f . This completes the proof. 

Proposition 3.2 Let 1 ≤ p < ∞. Every function f ∈ S can be approxi-

mated in Lp (Ω) by step functions.

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Proof:Let f = χE where E ⊂ Ω is a set of finite measure. Let ε > 0. Then,
by Lemma 2.1, there exists a set F , which is a finite disjoint union of boxes,
such that µ(E∆F ) < εp . Then
kχE − χF kpp = µ(E∆F ) < εp

and so kχE −χF kp < ε. Now, let f = kj=1 αj χEj , where the Ej are mutually
P
disjoint sets of finite measure and all the αj are non-zero. For each 1 ≤ j ≤ k,
choose Fj , a finite disjoint union of boxes such that kχEj − χFj kp < k|αε j | .
Consider the step function ϕ = kj=1 αj χFj . Then, by the triangle inequality,
P
it follows that kf − ϕkp < ε. This completes the proof. 
Proposition 3.3 Every step function defined on Ω can be approximated by
a function from Cc (Ω) in Lp (Ω), where 1 ≤ p < ∞.
Proof: Let f 6= 0 be a step function defined on Ω. Let ε > 0. By Corollary
2.1, there exists ϕ ∈ Cc (Ω) such that
 p
ε
µ({x ∈ Ω | ϕ(x) 6= f (x)}) <
2kf k∞
and such that (2.2) holds. Then
kϕ − f kpp ≤ 2p kf kp∞ µ({x ∈ Ω | ϕ(x) 6= χI (x)}) < εp .
Thus kϕ − f kp < ε. This completes the proof. 

Combining Propositions 3.1-3.3, we deduce the following result.

Theorem 3.1 Let Ω ⊂ RN be an open set and let 1 ≤ p < ∞. Then Cc (Ω)
is dense in Lp (Ω). 

4 Lusin’s theorem
Theorem 4.1 (Lusin) Let E ⊂ RN be a measurable set of finite measure.
Let f : E → R be a measurable function. Let ε > 0 be given. Then, there
exists ϕ ∈ Cc (RN ) such that
µ({x ∈ E | ϕ(x) 6= f (x)}) < ε.
Further, if f is bounded then we can ensure that
kϕk∞ ≤ kf k∞ .

8
Proof: For each positive integer n, define
En = {x ∈ E | |f (x)| ≤ n}.

Then En ↑ E. Choose N such that µ(E\EN ) < 3ε . Now definefe : RN → R

by 
f (x) if x ∈ EN ,
fe(x) =
0 if x ∈ RN \EN .
Since fe is bounded and since EN has finite measure, it follows that fe is inte-
grable on RN . Consequently, by Theorem 3.1, there exists a sequence {ϕn }
in Cc (RN ) such that ϕn → fe in L1 (RN ). Then, there exists a subsequence
{ϕnk } which converges to fe almost everywhere.
Now, since EN has finite measure, there exists F ⊂ EN such that µ(EN \F ) <
ε
3
and such that ϕnk → fe uniformly on F , by virtue of Egorov’s theorem.
Again, since F has finite measure, we can choose a compact set K ⊂ F such
that µ(F \K) < 3ε (cf. Lemma 2.2). Clearly µ(E\K) < ε.
Now, since ϕnk converges uniformly to fe on K, it follows that the restric-
tion of fe to K is continuous on K. But K ⊂ F ⊂ EN and so fe = f on K.
Thus, the restriction of f to K is continuous on K. We can then extend the
restriction of f to K, to all of RN by means of the Tietze extension theorem.
If f were bounded, this function, denoted g, will also satisfy
kgk∞ ≤ kf k∞ .
Finally, let ψ ∈ Cc (RN ) be such that 0 ≤ ψ ≤ 1 and such that ψ ≡ 1 on K
and set ϕ = ψg. Then ϕ ∈ Cc (RN ) and
{x ∈ E | ϕ(x) 6= f (x)} ⊂ E\K
and µ(E\K) < ε. This completes the proof. 

Remark 4.1 The extension theorem of Tietze is equivalent to Urysohn’s

lemma in a normal topological space (cf. Simmons [1]). 

References
[1] Simmons, G. F. Introduction to Topology and Modern Analysis,
McGraw-Hill, 1963.