## 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 $\int_E f_k\to\int_E f$.