proof of dominated convergence theoremProofOfDominatedConvergenceTheorem2003-03-07 07:37:582003-03-07 07:37:58Proofdominated convergence theorem0% this is the default PlanetMath preamble. as your knowledge
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% define commands hereIt is not difficult to prove that $f$ is measurable. In fact we can write
\[
f(x)=\sup_n \inf_{k\ge n} f_k(x)
\]
and we know that measurable functions are closed under the $\sup$ and $\inf$ operation.
Consider the sequence $g_n(x)=2\Phi(x) - \vert f(x)-f_n(x)\vert$.
Clearly $g_n$ are nonnegative functions since $f-f_n\le 2\Phi$.
So, applying Fatou's Lemma, we obtain
\begin{eqnarray*}
\lefteqn{\lim_{n\to\infty} \int_X \vert f-f_n\vert\, d\mu
\le \limsup_{n\to \infty} \int_X \vert f-f_n\vert\, d\mu}\\
& = & - \liminf_{n\to\infty} \int_X -\vert f-f_n\vert\, d\mu \\
& = & \int_X 2\Phi\, d\mu - \liminf_{n\to\infty}\int_X 2\Phi-\vert f-f_n\vert\,d\mu\\
& \le & \int_X 2\Phi\, d\mu - \int_X 2\Phi - \limsup_{n\to \infty}\vert f-f_n\vert\, d\mu \\
&= & \int_X 2\Phi\, d\mu - \int_X 2\Phi\, d\mu = 0.
\end{eqnarray*}