The Riesz-Fischer Theorem

# The Riesz-Fischer Theorem

Theorem 1 (The Riesz-Fischer Theorem): Let $(X, \mathfrak T, \mu)$ be a measure space and let $1 \leq p \leq \infty$. Then the Lebesgue space $L^p(X, \mathfrak T, \mu)$ is complete. |

*Recall that a Cauchy sequence converges if and only if it has a convergent subsequence. We use this result in proving the Riesz-Fischer theorem.*

**Proof:**Let $(f_n)_{n=1}^{\infty}$ be a sequence of functions in $L^p(X, \mathfrak T, \mu)$ that is Cauchy. Let $(\epsilon_n)_{n=1}^{\infty}$ be defined for each $n \in \mathbb{N}$ by:

\begin{align} \quad \epsilon_n = \frac{1}{2^n} \end{align}

- Observe that:

\begin{align} \quad \sum_{n=1}^{\infty} \epsilon_n = \sum_{n=1}^{\infty} \frac{1}{2^n} = 1 < \infty \end{align}

- Now since $(f_n)_{n=1}^{\infty}$ is a Cauchy sequence there exists a subsequence $(f_{n_k})_{k=1}^{\infty}$ of $(f_n)_{n=1}^{\infty}$ such that for all $k \in \mathbb{N}$ we have that:

\begin{align} \quad \| f_{n_k} - f_{n_{k+1}} \|_p \leq \frac{1}{4^n} = \epsilon_n^2 \end{align}

- By the results on the Lemma to the Riesz-Fischer Theorem (1 ≤ p < ∞) and the Lemma to the Riesz-Fischer Theorem (p = ∞) pages, there must exist a function $f \in L^p(X, \mathfrak T, \mu)$ such that:

\begin{align} \quad \lim_{k \to \infty} \| f_{n_k} - f \|_p = 0 \end{align}

- That is, $(f_{n_k})_{k=1}^{\infty}$ converges to $f$ in $L^p(X, \mathfrak T, \mu)$. But then every Cauchy sequence $(f_n)_{n=1}^{\infty}$ in $L^p(X, \mathfrak T, \mu)$ has a convergent subsequence $(f_{n_k})_{k=1}^{\infty}$ in $L^p(X, \mathfrak T, \mu)$. Therefore $(f_n)_{n=1}^{\infty}$ converges in $L^p(X, \mathfrak T, \mu)$. So $L^p(X, \mathfrak T, \mu)$ is complete. $\blacksquare$