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 $\int_E f_k\to\int_E f$ .

Author: mathtuition88

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