Proving Monotone Convergence Theorem with Fatou’s Lemma

The statement of the Monotone Convergence Theorem reads:

Suppose that the set $E$ is measurable, and let $(f_n)_{n=1}^\infty$ be a sequence of non-negative, measurable functions such that, for $x \in E$ ,

$0 \leq f_1(x) \leq f_2 (x) \leq \cdots$

Let $f$ be defined by $f_n(x) \rightarrow f(x)$ as $n \rightarrow \infty$ . Then,

$\int_E f_n d\mu \rightarrow \int_E f d\mu, \quad (n \rightarrow \infty)$

In this proof, Fatou’s lemma will be assumed.

Notice that $\lim_{n\rightarrow \infty} f_n = f$ implies that

$\int_E \lim_{n\rightarrow \infty } f_n = \int_E f$

and so by Fatou’s lemma,

$\int_E \lim_{n\rightarrow \infty } f_n = \int_E f \leq \liminf_{n\rightarrow \infty} \int_E f, \quad (1)$

for $\lim_{n \rightarrow \infty } f_n = f = \liminf_{n \rightarrow \infty} f_n, \quad (2)$

Now, since $f_{n} \leq f_{n+1} \leq \cdots \leq f$ , for every intger $n$ , and the $f_n's$ are bound below by 0, we have

$0 \leq \cdots \leq \int_E f_n \leq \cdots \leq \int_E f$ , for every $n \in \mathbb{N}$ . And so, taking the supremum for $k \geq n$ and passing to the limit gets

$\limsup_{n \rightarrow \infty} \int_E f_n \leq \int_E f, \quad (3)$ .

Now, combining (3) with (1) and (2) yields:

$\limsup_{n\rightarrow \infty } \int_E f_n \leq \int_E f = \int_E \liminf_{n\rightarrow \infty} f_n \leq \liminf_{n\rightarrow \infty} \int_E f_n$

hence

$\limsup_{n\rightarrow \infty} f_n d\mu \leq \int_E f d\mu \leq \liminf_{n\rightarrow \infty} \int_E f_n d\mu$ ,

therefore

$\lim_{n \rightarrow \infty} \int_E f_n d\mu = \int_E f d\mu$

which proves everything that was promised.

Proving Monotone Convergence Theorem with Fatou’s Lemma

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