User:Addem/riesz fischer

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
Equivalent to being a norm complete space, is the property that every absolutely summable sequence is summable. Let $\{f_n\}$ be an absolutely summable sequence, $f_n\in \map{\LL^p}\mu$, and define $\sum_{k=1}^\infty \norm {f_k} = B < \infty$. Also define $G_n = \sum_{k=1}^n|f_k|$ and $G=\sum_{k=1}^\infty |f_k|$.

It is clear that the conditions of the Monotone Convergence Theorem hold, so that $\int_X G^p = \lim_{n\to\infty}\int_X G_n^p$. By observing that

we can also say that $\int_X |G_n|^p \le B^p$ and therefore $\lim_{n\to\infty}\int_X |G_n|^p \le B^p$. Therefore we have that $\int_XG^p \le B^p < \infty$ which confirms that $G\in \map{\LL^p}\mu$.

Notice that in particular $G\in \map{\LL^p}\mu$ entails that $G<\infty$ a.e. Because absolute convergence entails conditional convergence, we then have $F=\sum_{k=1}^\infty f_k$ converges a.e.  Because $|F|\le G$ then $F\in \map{\LL^p}\mu$.

It only remains to show that $\sum_{k=1}^n f_k \to F$, which we can accomplish by the Lebesgue Dominated Convergence Theorem. Because $\left|F - \sum_{k=1}^n f_k\right|^p \le (2G)^p \in \map{\LL^1}\mu$ then the theorem applies, and we infer

$ \ds \norm{F-\sum_{k=1}^n f_k}_p^p = \int_X\left|F-\sum_{k=1}^n f_k\right|^p \to 0 $

Therefore by definition of convergence in $\map{\LL^p}\mu$ we have that $\sum_{k=1}^\infty f_k$ converges in $\map{\LL^p}\mu$. This shows that $\{f_k\}$ is summable, as desired.