Dominated Convergence Theorem - Proof of The Theorem

Proof of The Theorem

Lebesgue's dominated convergence theorem is a special case of the Fatou–Lebesgue theorem. Below, however, is a direct proof that uses Fatou’s lemma as the essential tool.

Since ƒ is the pointwise limit of the sequence (f_n) of measurable functions that is dominated by g, it is also measurable and dominated by g, hence it is integrable. Furthermore (these will be needed later),

$|f-f_n| \le |f| + |f_n| \leq 2g$

for all n and

$\limsup_{n\to\infty} |f-f_n| = 0.$

The second of these is trivially true (by the very definition of f). Using linearity and monotonicity of the Lebesgue integral,

$\biggl| \int_S{f\,d\mu} - \int_S{f_n\,d\mu} \biggr| = \biggl| \int_S{(f-f_n)\,d\mu} \biggr| \le \int_S{|f-f_n|\,d\mu} .$

By the reverse Fatou lemma (it is here that we use the fact that |f-f_n| is bounded above by an integrable function)

$\limsup_{n\to\infty} \int_S |f-f_n|\,d\mu \le \int_S \limsup_{n\to\infty} |f-f_n|\,d\mu = 0,$

which implies that the limit exists and vanishes i.e.

$\lim_{n\to\infty} \int_S |f-f_n|\,d\mu= 0.$

The theorem now follows.

If the assumptions hold only μ-almost everywhere, then there exists a μ-null set N ∈ Σ such that the functions ƒ_n1_N satisfy the assumptions everywhere on S. Then ƒ(x) is the pointwise limit of ƒ_n(x) for x ∈ S \ N and ƒ(x) = 0 for x ∈ N, hence ƒ is measurable. The values of the integrals are not influenced by this μ-null set N.

Read more about this topic: Dominated Convergence Theorem

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