Question 11.T.5: (Dominated Convergence Theorem) Let (fn) be a sequence of me...
(Dominated Convergence Theorem)
Let (f_{n}) be a sequence of measurable functions on Ω, and suppose f_{n} converges to f almost everywhere. If there is a function g ∈ \mathcal{L}^{1}(Ω) such that, for all n ∈ \mathbb{N},
|f_{n}(x)| ≤ g(x) a.e.,
then f ∈ \mathcal{L}^{1}(Ω) and
\underset{n→∞}{\lim} \int_{Ω} {f_{n}} dm = \int_{Ω} {f} dm . (11.22)
We know from Theorem 10.18 that f is measurable on Ω. Since |f| ≤ g (a.e.), we have
\int_{Ω}{|f |} dm ≤ \int_{Ω}{g} dm < ∞,
which implies f ∈ \mathcal{L}^{1}(Ω).
Let h_{n} = 2g − |f_{n} − f | , which is clearly in \mathcal{L}_{+}^{0}(Ω), and therefore satisfies
\int_{Ω} {\lim \inf h_{n}} dm ≤ \lim \inf \int_{Ω} {h_{n}} dm,
by Fatou’s lemma. Since h_{n} → 2g, we have
\int_{Ω} {2g} dm ≤ \lim \inf \left[\int_{Ω} {2g} dm − \int_{Ω} {|f_{n} − f |} dm \right]
= \int_{Ω} {2g} dm − \lim \sup \int_{Ω} {|f_{n} − f |} dm,
which implies
\lim \sup \int_{Ω} {|f_{n} − f |} dm ≤ 0.
But since \int_{Ω} {|f_{n} − f |} dm ≥ 0, it follows that \lim \inf \ \int_{Ω} {|f_{n} − f |} dm ≥ 0, and therefore
\lim \int_{Ω} {|f_{n} − f |} dm = 0
Now, by taking the limit as n → ∞ in the inequality
\left|\int_{Ω} {f_{n}} dm − \int_{Ω} {f} dm \right| ≤ \int_{Ω} {|f_{n} − f |} dm,
we obtain (11.22).