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.

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