$\displaystyle \text {-191-}$

## THEOREM A.1.2.

$\displaystyle \text {The Lebesgue space }L_p \text { is complete and, hence, is a Banach Space }$

### PROOF

$\displaystyle \text {Let } \lbrace x_n \rbrace \text{ be a Cauchy sequence in } L_p \text{ ; then } \forall k \enspace \exists \text{ an } N_k ,\text { s.t. if } n, m, \geq N_k \text{ we have}$ $\displaystyle \large \lVert x_n - x_m \rVert p \lt 2^{-k}$ $\displaystyle \text {There is no loss in generality if it is assumed that } N_k < N_{k+1} \enspace \forall k .$ $\displaystyle \text {Thus}$ $\displaystyle \large \lVert x_{N_k} - x_{N_{k+1}} \rVert p \lt 2^{-k} \qquad *$ $\displaystyle \text {We define}$ $\displaystyle \large y_n = \sum_{k=1}^n \lvert x_{N_k} - x_{N_{k-1}} \rvert \qquad (x_{N_0} = 0)$ $\displaystyle \text {Clearly } y_n \text { is positive and increasing with n so by applying Levi's Monotone Convergence Theorem;}$ $\displaystyle \large \Big( \int_\Omega \lvert y_n(w)\rvert ^p d\mu(w) \Big) ^{\frac 1 p} \mapsto \Big( \int_\Omega \lvert y(w)\rvert ^p d\mu(w) \Big) ^{\frac 1 p}$ $\displaystyle \text {where y denotes;}$ $\displaystyle \large y = \sum_{k=1}^\infty \lvert x_{N_k} - x_{N_{k-1}} \rvert$ $\displaystyle \text {Here}$ $\displaystyle \large \lVert y \rVert p \enspace = \enspace \lim\limits_{n \mapsto \infty} \lVert y_n \rVert p \enspace \leq \enspace \lim\limits_{n \mapsto \infty} \sum_{k=1}^\infty \lVert x_{N_k} - x_{N_{k-1}} \rVert p$ $\displaystyle \text {Using * we see that } \large \lVert y \rVert p \normalsize \text { is less than a finite constant, so that;}$ $\displaystyle \large \int_\Omega \lvert y_n(w)\rvert ^p d\mu(w) \lt \infty$ $\displaystyle \text {This gives } \large y = \sum_{k=1}^\infty \lvert x_{N_k} - x_{N_{k-1}} \rvert \lt \infty \normalsize \text { except on the null set } \large \mathscr{N}.$ $\displaystyle \text {It is immediately clear that } \large \sum_{k=1}^\infty ( x_{N_k} - x_{N_{k-1}} ) \normalsize \text { converges almost everywhere.}$
$\displaystyle \text {-192-}$ $\displaystyle \text {Letting}$ $\displaystyle \large x = \sum_{k=1}^\infty ( x_{N_k} - x_{N_{k-1}} ) \normalsize \text { we show } \large x \in L_p$ $\displaystyle \text {Clearly}$ $\displaystyle \large \lvert x \rvert = \lvert \sum_{k=1}^\infty ( x_{N_k} - x_{N_{k-1}} ) \rvert \enspace \leq \enspace \sum_{k=1}^\infty \lvert x_{N_k} - x_{N_{k-1}} \rvert = y \normalsize \quad \text { a.e. }$ $\displaystyle \text {Therefore}$ $\displaystyle \large ( \lVert x \rVert p ) ^p = \int_\Omega \lvert x(w)\rvert ^p d\mu(w) \enspace \leq \enspace \int_\Omega y^p d\mu = ( \lVert y \rVert p ) ^p \lt \infty$ $\displaystyle \text {which implies}$ $\displaystyle \large \lVert x \rVert p \lt \infty \normalsize \text { and } \large x \in L_p$ $\displaystyle \text {To show that } \large \lVert x_n - x \rVert p \mapsto 0 \normalsize \text { we first prove } \large \lVert x_{N_k} - x \rVert p \mapsto 0 \normalsize \text { as } \large k \mapsto \infty$ $\displaystyle \text {Now}$ $\displaystyle \large \lvert x_{N_k} - x_{N_j} \rvert \enspace \leq \enspace \sum_{i=k+1}^j \lvert x_{N_i} - x_{N_{i-1}} \rvert \enspace \lt y$ $\displaystyle \text {Therefore } \large \lvert x_{N_k} - x_{N_j} \rvert ^p \normalsize \text { is dominated by } \large y^p \normalsize \text { and we may apply Lebesgue's Dominance Convergence Theorem. }$ $\displaystyle \text {Letting } j \mapsto \infty \normalsize \text { we have, }$ $\displaystyle \large \lvert x_{N_k} - x \rvert ^p \enspace \leq y^p$ $\displaystyle \text {and}$ $\displaystyle \large \int \lim\limits_{k \mapsto \infty} \lvert x_{N_k} - x \rvert ^p d\mu \enspace = \enspace \lim\limits_{k \mapsto \infty} \int \lvert x_{N_k} - x \rvert ^p d\mu \enspace = \enspace \lim\limits_{k \mapsto \infty} ( \lVert x_{N_k} - x \rVert p) ^p \enspace = \enspace 0$ $\displaystyle \text {Now for fixed } \epsilon \gt 0, \enspace \exists \enspace N \normalsize \text { such that if } m,n \geq N;$ $\displaystyle \large \lVert x_n - x_m \rVert p \enspace \le \enspace \normalsize \frac {\epsilon}{2}$ $\displaystyle \text {So for } N_k, n \gt N$ $\displaystyle \large \lVert x_n - x_{N_k} \rVert p \enspace \le \enspace \normalsize \frac {\epsilon}{2} \qquad \qquad (1)$ $\displaystyle \text {and since }$ $\displaystyle \large \lVert x_{N_k} - x \rVert p \enspace \mapsto \enspace 0$ $\displaystyle \text {We can, by choosing a K sufficiently large obtain an } N_k \text { such that;}$ $\displaystyle \large \lVert x_{N_k} - x \rVert p \enspace \le \enspace \normalsize \frac {\epsilon}{2} \qquad \qquad (2)$
$\displaystyle \text {-193-}$ $\displaystyle \text {Substituting into the triangle inequality;}$ $\displaystyle \large \lVert x - x_n \rVert p \enspace \leq \enspace \lVert x - x_{N_k} \rVert + \lVert x_{N_k} - x_n \rVert \enspace \lt \enspace \normalsize \frac {\epsilon}{2} + \frac {\epsilon}{2} = \large \epsilon$ $\displaystyle \text {for some } n \geq N.$ $\displaystyle \text {Thus, } x_n \text { converges to } x \text { in } L_p , \enspace L_p \text { is complete and hence a Banach Space } \qquad \blacksquare$
$\displaystyle \text {We note here and use later that the particular space } L_2$ $\displaystyle \large \int_\Omega \lvert y_n(w)\rvert ^2 d\mu(w) \lt \infty$ $\displaystyle \text {with the 2-norm is a Banach Space.}$