Theorems for the Lebesgue Integral
Dung Le1
We now prove some convergence theorems for Lebesgue’s integral. The main question
is this. Let {fn } be a sequence of Lebesgue integrable functions on E and assume that fn
converges a.e. to f on E. Is f Lebesgue integrable and
Z
Z
lim
fn =
E
f
?
E
Let’s look at the following examples. Define the following functions on [0, 1].
fn (x) =
1
χ
(x) and gn (x) = nχ[0,1/n] (x).
x [1/n,1]
We easily see that fn → f with f (x) = 1/x if x > 0 and f (0) = 0. But f is not Lebesgue
integrable (why?)! Meanwhile, gn → g with g ≡ 0. But
Z
g = 0 6= 1 = lim
[0,1]
Z
gn .
[0,1]
We see that appropriate hypotheses need to be assumed to answer our question.
1
Convergence theorems
We start with the monotone convergence theorem.
Theorem 1.1 If fn isR a sequence
of nonnegative measurable functions defined on E and
R
fn % f on E then lim E fn = E f .
R
R
Proof: Obviously, E f ≥ lim E fn (why?). To prove the opposite inequality, for each
integer m choose an increasing sequence {sm,n } of simple functions so that sm,n % fm as
n → ∞. We then define
Sn (x) = sup{sk,n (x) : 1 ≤ k ≤ n}, for n ≥ 1.
Because sk,n ≤ sk,n+1 , we have (check it!)
Sn ≤ Sn+1 and sm,n ≤ Sn ≤ fn for 1 ≤ m ≤ n.
Letting n → ∞ in the last inequality, we get
fm ≤ lim Sn ≤ f
n→∞
for each m ∈ N . Let m → ∞ in the above inequality, we see that Sn % f . Therefore
Z
lim
Z
f.
Sn =
E
E
1
Department of Applied Mathematics, University of Texas at San Antonio, 6900 North Loop 1604 West,
San Antonio, TX 78249. Email: [email protected]
1
On the other hand, since sk,n ≤ fk ≤ fn for all k ≤ n, we have Sn ≤ fn . Hence
Z
Z
Sn ≤ lim
f = lim
E
Z
E
fn .
E
This completes the proof.
Next, we present the Fatou lemma, which may not seem very interesting but its
applications in different contexts will be seen apparent later.
Lemma 1.2 (Fatou’s lemma) If fn is a sequence of nonnegative measurable functions defined on E, then
Z
Z
fn ≥
lim inf
n→∞
lim inf fn .
E n→∞
E
Proof: The function f = lim inf n→∞ fn is nonnegative and measurable (why?). Set
hm = inf n≥m fn . Then fm ≥ hm and hm % f . By the above theorem, we have
Z
Z
f = lim
m→∞ E
E
But
Z
hm ≤ inf
Z
Z
n≥m E
E
hm .
fn =⇒ lim
m→∞ E
hm ≤ lim inf
Z
fn .
E
The proof is complete.
We also have (prove this!)
Lemma 1.3 If fn is a sequence of measurable functions defined on E and there is a
Lebesgue integrable function u such that fn ≤ u for all n, then
Z
fn ≤
lim sup
n→∞
E
Z
lim sup fn .
E
n→∞
The third convergence theorem is the Lebesgue dominated convergence theorem.
Theorem 1.4 Let {fn } be a sequence of measurable functions on E that converges to f a.e.
on E. Suppose that there exists a Lebesgue integrable function g on E such that |fn | ≤ g
for all n. Then f is Lebesgue integrable and
Z
lim
|fn − f | = 0.
E
In particular, we have lim
R
E
fn =
R
E
f.
Proof: Since |fn | ≤ |g| and fn → f we see that |f | ≤ g. Thus, |fn −f | ≤ |fn |+|f | ≤ 2g.
Because |fn − f | → 0 and 2g is Lebesgue integrable, Lemma 1.3 yields
lim sup |
Z
f−
E
Z
E
fn |≤ lim sup
Z
|fn − f | ≤
E
This gives the theorem.
2
Z
E
lim sup |fn − f | = 0.
2
Other theorems
We present here another version of the fundamental theorem we learn in calculus. Much
more general versions of this are available but they are out of the scope of this course.
Let’s start with the following
Lemma 2.1 Let f : [a, b] → IR be a measurable function. Assume that f is differentiable
a.e. on [a, b]. Then f 0 is measurable.
Proof: Extend the function f to the interval [a, b + 1] by setting f to be a constant on
[b, b + 1]. For each integer n, define the function fn : [a, b] → IR by
fn (x) = n(f (x +
1
) − f (x)).
n
Obviously, fn is measurable. Moreover, fn (x) converges to f 0 (x) (why?) whenever f 0 (x) is
defined. Hence, fn converges to f 0 a.e. and, therefore, f 0 is measurable.
The proof of the following simple lemma is left as a homework.
Lemma 2.2 Let f : [a, b] → IR be a measurable function. If f is continuous at x ∈ [a, b)
then
1
Z
x+ n
lim n
n→∞
f = f (x).
x
Lemma 2.3 Let a ∈ IR and E ⊂ IR. Let f : E (E + a) → IR be a measurable function.
We have
Z
Z
f (x + a) =
f.
S
E
E+a
Proof: Using the translation invariant property of Lebesgue measure to prove this for
simple functions. The general case follows by limit argument (your homework!).
We then have the following version of the fundamental theorem of calculus.
Theorem 2.4 Let f : [a, b] → IR be a differentiable function. If f 0 is bounded, then f 0 is
Lebesgue integrable on [a, b]. Moreover, for any c ∈ [a, b], we have
Z
c
f 0 (x)dx = f (c) − f (a).
a
Proof: Let M = sup[a,b] |f 0 (x)|. We assume first that c < b. For n sufficiently large,
we consider the function fn defined on [a, c] by
fn (x) = n(f (x +
1
) − f (x)).
n
By the Mean Value Theorem, for each such integer n and each x ∈ [a, c], there exists
znx ∈ (x, x + 1/n) such that
fn (x) =
f (x + n1 ) − f (x)
1
n
3
= f 0 (znx ) ≤ M.
Thus, fn ≤ M , for all n. Since fn → f 0 on [a, c], Lebesgue dominated convergence theorem
gives
Z
Z
c
c
f 0 = lim
fn .
n→∞ a
a
Using the above lemmas and the continuity of f at a, c, we have
Z
c
f0 =
a
Z
c
Z
lim
n→∞ a
Z
=
lim n
n→∞
c
1
)−
!n
Z
c
a
f (x) −
Z
c
f (x)
a+1/n
c+1/n
Z
n→∞
a
Z
c+1/n
f (x) −
= lim n
n→∞
a
f (x) − lim n
f (x)
f (x +
n→∞
c+1/n
lim n
n→∞
Z
=
c
fn = lim n
c
Z
!
a+1/n
f (x)
a
a+1/n
f (x) = f (c) − f (a).
a
This completes the proof for c < b. For c = b, we can extend the function f to a, b + 1]
by defining f (x) = f 0 (b)(x − b) + f (b) for x ∈ (b, b + 1] and reduce this case to the one
discussed before.
Exercises
1. Show that f is Lebesgue integrable iff f is Lebesgue integrable.
√
2. Let f (x) = 1/ x for x > 0 and f (0) = 0. Show that f is Lebesgue integrable.
3. Is f (x) = 1/x Lebesgue integrable on (0, 1)?
4. RLet f : E → IR be a measurable function. Assume that f 6= 0 a.e. on E and that
E f = 0. Show that f = 0 a.e. on E.
5. Let f : [0, 1] → IR be a Lebesgue integrable function. Compute limn→∞
6. For a, b ∈ IR, compute limn→∞
Rb
a
sinn (x)dx and limn→∞
Rb
a
R1
0
xn f (x).
cosn (x)dx.
7. Let {fn } be a sequence of nonnegative measurable functions on E. Show that
Z X
∞
fn =
E n=1
∞ Z
X
fn .
n=1 E
8. For each integer n and x > 0, define fn (x) =
√
n
x/x. Show that
a) fn → f pointwise with f (x) = 1/x on the set x > 0.
b) {fn } is monotone on (0, 1) and (1, ∞).
√
c) For any z > 0, ln(z) = limn→∞ n( n z − 1).
9. Let f : [a, b] → IR be Lebesgue integrable. Prove that the function F (x) =
uniformly continuous on [a, b].
4
Rx
a
f is