proof of dominated convergence theorem

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

$$f(x)=\underset{n}{sup}\underset{k\ge n}{inf}{f}_k(x)$$

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

Consider the sequence $g_n(x)=2\mathrm{\Phi }(x)-|f(x)-f_n(x)|$. Clearly $g_n$ are nonnegative functions since $f-f_n\le 2\mathrm{\Phi }$. So, applying Fatou’s Lemma, we obtain

$\underset{n\to \mathrm{\infty }}{lim}{\displaystyle {\int }_X}|f-f_n|𝑑\mu \le \underset{n\to \mathrm{\infty }}{lim\; sup}{\displaystyle {\int }_X}|f-f_n|𝑑\mu$

$=$ $-\underset{n\to \mathrm{\infty }}{lim\; inf}{\displaystyle {\int }_X}-|f-f_n|d\mu$

$=$ ${\displaystyle {\int }_X}2\mathrm{\Phi }𝑑\mu -\underset{n\to \mathrm{\infty }}{lim\; inf}{\displaystyle {\int }_X}2\mathrm{\Phi }-|f-f_n|d\mu$

$\le$ ${\displaystyle {\int }_X}2\mathrm{\Phi }𝑑\mu -{\displaystyle {\int }_X}2\mathrm{\Phi }-\underset{n\to \mathrm{\infty }}{lim\; sup}|f-f_n|d\mu$

$=$ ${\displaystyle {\int }_X}2\mathrm{\Phi }𝑑\mu -{\displaystyle {\int }_X}2\mathrm{\Phi }𝑑\mu =0.$

Title	proof of dominated convergence theorem
Canonical name	ProofOfDominatedConvergenceTheorem
Date of creation	2013-03-22 13:30:02
Last modified on	2013-03-22 13:30:02
Owner	paolini (1187)
Last modified by	paolini (1187)
Numerical id	4
Author	paolini (1187)
Entry type	Proof
Classification	msc 28A20
Related topic	SecondProofOfDominatedConvergenceTheorem
Related topic	SecondProofOfDominatedConvergenceTheorem2