# proof of dominated convergence theorem

It is not difficult to prove that $f$ is measurable. In fact we can write

 $f(x)=\sup_{n}\inf_{k\geq n}f_{k}(x)$

and we know that measurable functions  are closed under the $\sup$ and $\inf$ operation.

Consider the sequence $g_{n}(x)=2\Phi(x)-|f(x)-f_{n}(x)|$. Clearly $g_{n}$ are nonnegative functions since $f-f_{n}\leq 2\Phi$. So, applying Fatou’s Lemma, we obtain

 $\displaystyle\lim_{n\to\infty}\int_{X}|f-f_{n}|\,d\mu\leq\limsup_{n\to\infty}% \int_{X}|f-f_{n}|\,d\mu$ $\displaystyle=$ $\displaystyle-\liminf_{n\to\infty}\int_{X}-|f-f_{n}|\,d\mu$ $\displaystyle=$ $\displaystyle\int_{X}2\Phi\,d\mu-\liminf_{n\to\infty}\int_{X}2\Phi-|f-f_{n}|\,d\mu$ $\displaystyle\leq$ $\displaystyle\int_{X}2\Phi\,d\mu-\int_{X}2\Phi-\limsup_{n\to\infty}|f-f_{n}|\,d\mu$ $\displaystyle=$ $\displaystyle\int_{X}2\Phi\,d\mu-\int_{X}2\Phi\,d\mu=0.$
Title proof of dominated convergence theorem  ProofOfDominatedConvergenceTheorem 2013-03-22 13:30:02 2013-03-22 13:30:02 paolini (1187) paolini (1187) 4 paolini (1187) Proof msc 28A20 SecondProofOfDominatedConvergenceTheorem SecondProofOfDominatedConvergenceTheorem2