# Mathematics Prelims

## July 20, 2008

### The Monotone Convergence Theorem

Filed under: Analysis,Measure Theory — cjohnson @ 4:59 pm

If $(f_n : E \to \mathbb{R})_{n \in \mathbb{N}}$ is a sequence of non-negative measurable functions and $(f_n(x))_{n \in \mathbb{N}}$ increases monotonically to $f(x)$ for each $x \in E$, $f_n \nearrow f$ pointwise (this can actually be relaxed to just having $f_n \nearrow f$ a.e.), then

$\displaystyle \lim_{n \to \infty} \int_E f_n(x) \, dm = \int_E f \, dm$.

Proof:

As $f_n \nearrow f$ a.e., we have that $f_n \leq f$ a.e. for each $n \in \mathbb{N}$, and so $\int_E f_n \, dm \leq \int_E f \, dm$ for each $n$.  Now also notice that since $f_n \leq f_{n + 1}$, that $\int_E f_n \, dm$ forms a bounded monotonic sequence sequence of real numbers, so it must converge.  So we have

$\displaystyle \lim_{n \to \infty} \int_E f_n \, dm \leq \int_E f \, dm = \int_E \lim_{n \to \infty} f \, dm$

By Fatou’s lemma we have that

$\displaystyle \int_E f \, dm = \int_E \lim_{n \to \infty} f_n \, dm = \int_E \liminf_{n \to \infty} f_n \, dm$

$\displaystyle \qquad \leq \liminf_{n \to \infty} \int_E f_n \, dm = \lim_{n \to \infty} \int_E f_n \, dm$

Since we have inequalities both ways, we must have equality.