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 (fn) 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-fn| 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 ƒn1N satisfy the assumptions everywhere on S. Then ƒ(x) is the pointwise limit of ƒn(x) for xS \ N and ƒ(x) = 0 for xN, 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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