# Proving DCT from Fatou's Lemma

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 $$n \rightarrow \infty$$,

$$\vert f(x) \vert \leq g(x)$$ ; $$g(x)$$ 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?

• The condition $|f| \leq g$ a.e. implies $g \geq 0$ a.e. – Will M. Dec 19 '18 at 23:22
• omg I cannot believe I missed that, thanks!! – 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 $$|f_n-f|$$