Riesz-Fischer Theorem

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $p \in \R$, $p \ge 1$.

The Lebesgue $p$-space $\map {\LL^p} \mu$, endowed with the $p$-norm, is a complete metric space.

Proof
From Norm Complete iff Absolutely Summable Entails Summable, every absolutely summable sequence is summable.

Let $\sequence {f_n}$ be an absolutely summable sequence:
 * $f_n \in \map {\LL^p} \mu$

Define:
 * $\ds \sum_{k \mathop = 1}^\infty \norm {f_k} =: B < \infty$

Also define:
 * $\ds G_n := \sum_{k \mathop = 1}^n \size {f_k}$

and:
 * $\ds G = \sum_{k \mathop = 1}^\infty \size {f_k}$

It is clear that the conditions of the Monotone Convergence Theorem (Measure Theory) hold, so that:
 * $\ds \int_X G^p = \lim_{n \mathop \to \infty} \int_X G_n^p$

By observing that:

we can also say that:
 * $\ds \int_X \size {G_n}^p \le B^p$

and therefore:
 * $\ds \lim_{n \mathop \to \infty} \int_X \size {G_n}^p \le B^p$

Therefore we have that:
 * $\ds \int_X G^p \le B^p < \infty$

This confirms:
 * $G \in \map {\LL^p} \mu$

In particular:
 * $G \in \map{\LL^p} \mu$

entails that:
 * $G < \infty$ a.e.

Because absolute convergence entails conditional convergence:
 * $\ds F = \sum_{k \mathop = 1}^\infty f_k$

converges a.e.

Because $\size F \le G$:
 * $F \in \map {\LL^p} \mu$

It only remains to show that:
 * $\ds \sum_{k \mathop = 1}^n f_k \to F$

which we can accomplish by Lebesgue's Dominated Convergence Theorem.

Because $\ds \size {F - \sum_{k \mathop = 1}^n f_k}^p \le (2G)^p \in \map{\LL^1}\mu$, the theorem applies.

We infer:


 * $\ds \norm {F - \sum_{k \mathop = 1}^n f_k}_p^p = \int_X \size {F - \sum_{k \mathop = 1}^n f_k}^p \to 0$

Therefore by definition of convergence in $\map{\LL^p}\mu$ we have that $\ds \sum_{k \mathop = 1}^\infty f_k$ converges in $\map{\LL^p}\mu$.

This shows that $\sequence {f_k}$ is summable, as we were to prove.