Theorems for the Lebesgue Integral
Dung Le
1
We now prove some convergence theorems for Lebesgue’s integral.The main question
is this.Let {f
n
} be a sequence of Lebesgue integrable functions on E and assume that f
n
converges a.e.to f on E.Is f Lebesgue integrable and
lim
E
f
n
=
E
f?
Let’s look at the following examples.Deﬁne the following functions on [0,1].
f
n
(x) =
1x
χ
[1/n,1]
(x) and g
n
(x) = nχ
[0,1/n]
(x).
We easily see that f
n
→f with f(x) = 1/x if x > 0 and f(0) = 0.But f is not Lebesgue
integrable (why?)!Meanwhile,g
n
→g with g ≡ 0.But
[0,1]
g = 0 = 1 = lim
[0,1]
g
n
.
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 f
n
is a sequence of nonnegative measurable functions deﬁned on E and
f
n
f on E then lim
E
f
n
=
E
f.
Proof:Obviously,
E
f ≥ lim
E
f
n
(why?).To prove the opposite inequality,for each
integer m choose an increasing sequence {s
m,n
} of simple functions so that s
m,n
f
m
as
n →∞.We then deﬁne
S
n
(x) = sup{s
k,n
(x):1 ≤ k ≤ n},for n ≥ 1.
Because s
k,n
≤ s
k,n+1
,we have (check it!)
S
n
≤ S
n+1
and s
m,n
≤ S
n
≤ f
n
for 1 ≤ m≤ n.
Letting n →∞in the last inequality,we get
f
m
≤ lim
n→∞
S
n
≤ f
for each m∈ N.Let m→∞in the above inequality,we see that S
n
f.Therefore
lim
E
S
n
=
E
f.1
Department of Applied Mathematics,University of Texas at San Antonio,6900 North Loop 1604 West,
San Antonio,TX 78249.Email:dle@math.utsa.edu
1
On the other hand,since s
k,n
≤ f
k
≤ f
n
for all k ≤ n,we have S
n
≤ f
n
.Hence
E
f = lim
E
S
n
≤ lim
E
f
n
.
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 f
n
is a sequence of nonnegative measurable functions de
ﬁned on E,then
liminf
n→∞
E
f
n
≥
E
liminf
n→∞
f
n
.
Proof:The function f = liminf
n→∞
f
n
is nonnegative and measurable (why?).Set
h
m
= inf
n≥m
f
n
.Then f
m
≥ h
m
and h
m
f.By the above theorem,we have
E
f = lim
m→∞
E
h
m
.
But
E
h
m
≤ inf
n≥m
E
f
n
=⇒ lim
m→∞
E
h
m
≤ liminf
E
f
n
.
The proof is complete.We also have (prove this!)
Lemma 1.3 If f
n
is a sequence of measurable functions deﬁned on E and there is a
Lebesgue integrable function u such that f
n
≤ u for all n,then
limsup
n→∞
E
f
n
≤
E
limsup
n→∞
f
n
.
The third convergence theorem is the Lebesgue dominated convergence theorem.
Theorem 1.4 Let {f
n
} 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 f
n
 ≤ g
for all n.Then f is Lebesgue integrable and
lim
E
f
n
−f = 0.
In particular,we have lim
E
f
n
=
E
f.
Proof:Since f
n
 ≤ g and f
n
→f we see that f ≤ g.Thus,f
n
−f ≤ f
n
+f ≤ 2g.
Because f
n
−f →0 and 2g is Lebesgue integrable,Lemma 1.3 yields
limsup 
E
f −
E
f
n
≤ limsup
E
f
n
−f ≤
E
limsupf
n
−f = 0.
This gives the theorem.2
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
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 f
n
:[a,b] →IR by
f
n
(x) = n(f(x +
1n
) −f(x)).
Obviously,f
n
is measurable.Moreover,f
n
(x) converges to f
(x) (why?) whenever f
(x) is
deﬁned.Hence,f
n
converges to f
a.e.and,therefore,f
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
lim
n→∞
n
x+
1n
x
f = f(x).
Lemma 2.3 Let a ∈ IR and E ⊂ IR.Let f:E
(E +a) → IR be a measurable function.
We have
E
f(x +a) =
E+a
f.
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
is bounded,then f
is
Lebesgue integrable on [a,b].Moreover,for any c ∈ [a,b],we have
c
a
f
(x)dx = f(c) −f(a).
Proof:Let M = sup
[a,b]
f
(x).We assume ﬁrst that c < b.For n suﬃciently large,
we consider the function f
n
deﬁned on [a,c] by
f
n
(x) = n(f(x +
1n
) −f(x)).
By the Mean Value Theorem,for each such integer n and each x ∈ [a,c],there exists
z
x
n
∈ (x,x +1/n) such that
f
n
(x) =
f(x +
1 n
) −f(x)1n
= f
(z
x
n
) ≤ M.
3
Thus,f
n
≤ M,for all n.Since f
n
→f
on [a,c],Lebesgue dominated convergence theorem
gives
c
a
f
= lim
n→∞
c
a
f
n
.
Using the above lemmas and the continuity of f at a,c,we have
c
a
f
= lim
n→∞
c
a
f
n
= lim
n→∞
n
c
a
f(x +
1n
) −
c
a
f(x)
= lim
n→∞
n
c+1/n
a+1/n
f(x) −
c
a
f(x)
= lim
n→∞
n
c+1/n
c
f(x) −
a+1/n
a
f(x)
= lim
n→∞
n
c+1/n
c
f(x) − lim
n→∞
n
a+1/n
a
f(x) = f(c) −f(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
(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.Let f:E → IR be a measurable function.Assume that f = 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 lim
n→∞
1
0
x
n
f(x).
6.For a,b ∈ IR,compute lim
n→∞
b
a
sin
n
(x)dx and lim
n→∞
b
a
cos
n
(x)dx.
7.Let {f
n
} be a sequence of nonnegative measurable functions on E.Show that
E
∞
n=1
f
n
=
∞
n=1
E
f
n
.
8.For each integer n and x > 0,deﬁne f
n
(x) =
n
√x/x.Show that
a) f
n
→f pointwise with f(x) = 1/x on the set x > 0.
b) {f
n
} is monotone on (0,1) and (1,∞).
c) For any z > 0,ln(z) = lim
n→∞
n(
n
√ z −1).
9.Let f:[a,b] → IR be Lebesgue integrable.Prove that the function F(x) =
x
a
f is
uniformly continuous on [a,b].
4
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