📂Measure Theory

Proof of the Monotone Convergence Theorem

Theorem ¹

Let a sequence $\left\{ f_{n} \right\}$ of measurable functions with nonnegative values satisfy $f_{n} \nearrow f$. Then $$ \lim_{n \to \infty} \int_{E} f_{n} dm = \int_{E} f dm $$

Explanation

$f_{n} \nearrow f$ means that for all $x$ we have $f_{n}(x) \le f_{n+1} (x)$ and $\displaystyle \lim_{n \to \infty} f_{n} = f$. Since the formula is so simple, knowing this theorem means knowing its ‘conditions’ precisely. As for usefulness, it means that the limit can move freely in and out of the integral, which goes without saying.

Proof

Since $f_{n} \le f$, $$ \limsup_{n \to \infty} \int_{E} f_{n} dm \le \int_{E} f dm $$

Fatou’s lemma: For a sequence $\left\{ f_{n} \right\}$ of measurable functions with nonnegative values, $$\displaystyle \int_{E} \left( \liminf_{n \to \infty} f_{n} \right) dm \le \liminf_{n \to \infty} \int_{E} f_{n} dm $$

By Fatou’s lemma and the properties of the limit infimum, $\displaystyle \int_{E} f dm \le \liminf_{n \to \infty} \int_{E} f_{n} dm$ holds, and rearranging gives $$\displaystyle \limsup_{n \to \infty} \int_{E} f_{n} dm \le \int_{E} f dm \le \liminf_{n \to \infty} \int_{E} f_{n} dm $$ But obviously $\displaystyle \liminf_{n \to \infty} \int_{E} f_{n} dm \le \limsup_{n \to \infty} \int_{E} f_{n} dm$, so it must be that $$\displaystyle \limsup_{n \to \infty} \int_{E} f_{n} dm = \int_{E} f dm = \liminf_{n \to \infty} \int_{E} f_{n} dm$$

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Corollary

Let a sequence $\left\{ f_{n} \right\}$ of measurable functions with nonnegative values satisfy $f_{n} \nearrow f$ almost everywhere. Then $$\lim_{n \to \infty} \int_{E} f_{n} dm = \int_{E} f dm$$ and, in particular, $$ \int \sum_{n=1}^{\infty} f_{n} dm = \sum_{n=1}^{\infty} \int f_{n} dm$$