Lebesgue’s Dominated Convergence Theorem for Convergence in Measure

Lebesgue’s Dominated Convergence Theorem for Convergence in Measure

If \{f_k\} satisfies f_k\xrightarrow{m}f on E and |f_k|\leq\phi\in L(E), then f\in L(E) and \int_E f_k\to\int_E f.

Proof

Let \{f_{k_j}\} be any subsequence of \{f_k\}. Then f_{k_j}\xrightarrow{m}f on E. Thus there is a subsequence f_{k_{j_l}}\to f a.e.\ in E. Clearly |f_{k_{j_l}}|\leq\phi\in L(E).

By the usual Lebesgue’s DCT, f\in L(E) and \int_E f_{k_{j_l}}\to\int_E f.

Since every subsequence of \{\int_E f_k\} has a further subsequence that converges to \int_E f, we have \int_E f_k\to\int_E f.

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