20.8 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 definition.
Lemma 20.8.1 Let
be a sequence in
limn→∞an exists if and only
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 20.8.2 (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 Theorem 19.1.2. Since |f|≤ g, it follows that
By Fatou’s lemma (Theorem 20.5.1),
This proves the theorem by Lemma 20.8.1
because the limsup and liminf are equal. ■
Corollary 20.8.3 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 20.8.4 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