6.8 The Dominated Convergence Theorem
One of the major theorems in this theory is the dominated convergence theorem. Before presenting it,
here is a technical lemma about limsup and liminf which is really pretty obvious from the
Lemma 6.8.1 Let
be a sequence in
limn→∞an exists if and only if
and in this case, the limit equals the common value of these two numbers.
Proof: Suppose first limn→∞an = a ∈ ℝ. Then, letting ε > 0 be given, an ∈
large enough, say n ≥ N.
Therefore, both inf
are contained in
n ≥ N.
It follows limsupn→∞an
and liminf n→∞an
are both in
Since ε is arbitrary, the two must be equal and they both must equal a. Next suppose limn→∞an = ∞.
Then if l ∈ ℝ, there exists N such that for n ≥ N,
and therefore, for such n,
and this shows, since l is arbitrary that
The case for −∞ is similar.
Conversely, suppose liminf n→∞an = limsupn→∞an = a. Suppose first that a ∈ ℝ. Then, letting ε > 0
be given, there exists N such that if n ≥ N,
therefore, if k,m > N, and ak > am,
is a Cauchy sequence. Therefore, it converges to
a ∈ ℝ
, and as in the first part, the
liminf and limsup both equal a.
If liminf n→∞an
then given l ∈ ℝ
, there exists N
such that for n ≥ N,
Therefore, limn→∞an = ∞. The case for −∞ is similar. ■
Here is the dominated convergence theorem.
Theorem 6.8.2 (Dominated Convergence theorem) Let fn ∈ L1(Ω) and suppose
and there exists a measurable function g, with values in
Then f ∈ L1
Proof: f is measurable by Corollary 5.1.3 applied to real and imaginary parts. Since |f|≤ g, it follows
By Fatou’s lemma (Theorem 6.5.1),
This proves the theorem by Lemma 6.8.1
because the limsup and liminf are equal. ■
Corollary 6.8.3 Suppose fn ∈ L1
. Suppose also there exist measurable
functions, gn, g with values in
μ a.e. and both
gndμ and ∫
gdμ are finite. Also suppose
Proof: It is just like the above. This time g + gn −
0 and so by Fatou’s lemma,
and so −
Definition 6.8.4 Let E be a measurable subset of Ω.
If L1(E) is written, the σ algebra is defined as
and the measure is μ restricted to this smaller σ algebra. Clearly, if f ∈ L1(Ω), then
and if f ∈ L1(E), then letting
be the 0 extension of
off of E
, it follows
What about something ordinary, the integral of a continuous function?
Theorem 6.8.5 Let f be continuous on
where the integral on the left is the usual Riemann integral and the integral on the right is the Lebesgue
Proof: From Theorems 5.7.1 and 5.10.1 fX
is Lebesgue measurable. Assume for the sake of
simplicity that f
0. If not, apply what is about to be shown to f+
. Let sn
be a step
function and let this converge uniformly to
= 0 for
. For example,
thanks to Theorem 5.7.1 which gives the measure of intervals. Then one can apply the definition of the
Riemann integral to obtain the left side converging to ∫
is continuous and
is nothing more than a left sum. Then apply the dominated convergence theorem on the right
to obtain the claim of the theorem. Indeed f
is bounded and so there exists M ≥ f
0 for all x ∈
This shows that for reasonable functions, there is nothing new in the Lebesgue integral.
The big difference is that now you have limit theorems which may be applied and you can
integrate more functions. In fact, every Riemann integrable function on an interval is Lebesgue