Forgive me, I am new to measure theory. I am trying to prove the Dominated Convergence Theorem by assuming Fatou's Lemma, here is what I have so far.

Fatou's Lemma:

Let $\{f_n\} $ be a sequence of nonnegative measurable functions s.t.

$f_n \rightarrow f$ a.e. x as $n \rightarrow \infty$,

then $\int f \leq $ lim$_{n \rightarrow \infty}$ inf $\int f_n$.

The DCT states:

Let $\{f_n\}$ be sequence of measurable functions s.t.

$f_n \rightarrow f$ a.e. x as $n \rightarrow \infty$,

if in addition we have

$\vert f(x) \vert \leq g(x)$ ; $g(x)$ integrable,

then $\int \vert f_n - f \vert \rightarrow 0$, as $n \rightarrow \infty$.

I know one way to prove this is you can define a set of elements bounded above by integer values so the functions are supported on a set of finite measure allowing the use of the bounded convergence theorem.

Now Fatou's Lemma takes into consideration the nonnegative functions, something I cannot assume with the DCT, but since the $f_n$ are all bounded above by an integrable function $g(x)$ could I rewrite $g(x)$ as its decomposition into $g^+ - g^-$?

My intuition says the result will "pop out" if I had non negativity? OR am I missing something else here?

  • 1
    $\begingroup$ The condition $|f| \leq g$ a.e. implies $g \geq 0$ a.e. $\endgroup$ – Will M. Dec 19 '18 at 23:22
  • $\begingroup$ omg I cannot believe I missed that, thanks!! $\endgroup$ – Hossien Sahebjame Dec 22 '18 at 20:53

$\int [g-f]=\int \lim \inf [g-f_n] \leq \lim \inf \int [g-f_n]$ which gives $\int g -\int f \leq \int g -\lim \sup \int f_n$ so $\lim \sup \int f_n \leq \int f$. Now replace $f$ by $-f$ and $f_n$ by $-f_n$ to get $\lim \inf \int f_n \geq \int f$. Hence $\int f_n \to \int f$. To get $\int |f_n-f| \to 0$ you simply have to replace $f$ by $0$, $g$ by $2g$ and $f_n$ by $|f_n-f|$

  • $\begingroup$ ok I finally understand it haha thanks so much man!!! $\endgroup$ – Hossien Sahebjame Dec 23 '18 at 3:17

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