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 6.8.1Let
{an}
be a sequence in
[− ∞,∞ ]
. Then lim_{n→∞}a_{n}exists if and only if
lim in→nf∞ an = lim ns→up∞ an
and in this case, the limit equals the common value of these two numbers.
Proof:Suppose first lim_{n→∞}a_{n} = a ∈ ℝ. Then, letting ε > 0 be given, a_{n}∈
(a− ε,a +ε)
for all n
large enough, say n ≥ N. Therefore, both inf
{ak : k ≥ n}
and sup
{ak : k ≥ n}
are contained in
[a − ε,a + ε]
whenever n ≥ N. It follows limsup_{n→∞}a_{n} and liminf _{n→∞}a_{n} are both in
[a − ε,a + ε]
,
showing
| |
||lim inf an − lim sup an||< 2ε.
| n→ ∞ n→∞ |
Since ε is arbitrary, the two must be equal and they both must equal a. Next suppose lim_{n→∞}a_{n} = ∞.
Then if l ∈ ℝ, there exists N such that for n ≥ N,
l ≤ an
and therefore, for such n,
l ≤ inf{ak : k ≥ n} ≤ sup {ak : k ≥ n}
and this shows, since l is arbitrary that
lim inf an = lim sup an = ∞.
n→∞ n→∞
The case for −∞ is similar.
Conversely, suppose liminf _{n→∞}a_{n} = limsup_{n→∞}a_{n} = a. Suppose first that a ∈ ℝ. Then, letting ε > 0
be given, there exists N such that if n ≥ N,
sup{ak : k ≥ n }− inf{ak : k ≥ n} < ε
therefore, if k,m > N, and a_{k}> a_{m},
|ak − am | = ak − am ≤ sup{ak : k ≥ n} − inf{ak : k ≥ n} < ε
showing that
{a }
n
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→∞}a_{n} = limsup_{n→∞}a_{n} = ∞, then given l ∈ ℝ, there exists N
such that for n ≥ N,
inf an > l.
n>N
Therefore, lim_{n→∞}a_{n} = ∞. The case for −∞ is similar. ■
Here is the dominated convergence theorem.
Theorem 6.8.2(Dominated Convergence theorem)Let f_{n}∈ L^{1}(Ω) and suppose
f(ω) = lim fn(ω ),
n→ ∞
and there exists a measurable function g, with values in
[0,∞],^{1}such that
∫
|fn(ω )| ≤ g(ω ) and g(ω)dμ < ∞.
Then f ∈ L^{1}
(Ω)
and
∫ |∫ ∫ |
0 = lim |f − f|dμ = lim || fdμ − f dμ||
n→∞ n n→∞ | n |
Proof: f is measurable by Corollary 5.1.3 applied to real and imaginary parts. Since |f|≤ g, it follows
that
is Lebesgue measurable. Assume for the sake of
simplicity that f
(x)
≥ 0. If not, apply what is about to be shown to f^{+} and f^{−}. Let s_{n}
(x)
be a step
function and let this converge uniformly to f
(x)
on
[a,b]
with s_{n}
(x)
= 0 for x
∈∕
[a,b]
. For example,
let
∑n
sn(x) ≡ f (xj−1)X [xj−1,xj)(x)
j=1
Then
∫ b ∫
sn (x)dx = sndm
a [a,b]
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 ∫_{a}^{b}f
(x )
dx because f is continuous and
∫_{a}^{b}s_{n}
(x)
dx 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
(x)
≥ 0 for all x ∈
[a,b]
.
■
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
integrable.