Riesz-Fischer completeness of $L^p$ for $1\le p\le \mathrm{\infty }$

Assume $1\le p<\mathrm{\infty }$ and let $(u_n)$ be Cauchy in $L^p(\mu )$. Choose by least indices a strictly increasing sequence $(n_k)$ such that [L2, L3, L4, L5, given, choose] $\mathrm{\parallel }{u}_{n_{k+1}}-u_{n_k}{\mathrm{\parallel }}_p<2^{-k}(k\ge 0)\mathrm{.}$ Use countable choice (the countable instance of AC) to choose measurable representatives $f_k$ of $u_{n_k}$, and put $h_k:=\mathrm{\mid }{f}_{k+1}-f_k\mathrm{\mid }$. Then each $h_k$ is measurable, belongs to ${\mathcal{L}}^p(\mu )$, and satisfies $\mathrm{\parallel }{h}_k{\mathrm{\parallel }}_p<2^{-k}\mathrm{.}$ If $g_m:={\sum }_{k<m}{h}_k$, then $\mathrm{\parallel }{g}_m{\mathrm{\parallel }}_p\le {\sum }_{k<m}\mathrm{\parallel }{h}_k{\mathrm{\parallel }}_p\le {\sum }_{k=0}^{\mathrm{\infty }}{2}^{-k}<\mathrm{\infty },$ so each $g_m$ lies in ${\mathcal{L}}^p(\mu )$. Monotone convergence then gives a measurable pointwise limit $g:={\sum }_{k=0}^{\mathrm{\infty }}{h}_k$ with $g\in {\mathcal{L}}^p(\mu )$.

Assume now $p=\mathrm{\infty }$ and let $(u_n)$ be Cauchy in $L^{\mathrm{\infty }}(\mu )$. Choose least indices $n_k$ with [L4, L6, L7, given, choose] $\mathrm{\parallel }{u}_{n_{k+1}}-u_{n_k}{\mathrm{\parallel }}_{\mathrm{\infty }}<2^{-k}\mathrm{.}$ Use countable choice (the countable instance of AC) to choose measurable representatives $f_k$ of $u_{n_k}$. By [L7], the explicitly defined measurable set $E_k:=\{x:\mathrm{\mid }{f}_{k+1}(x)-f_k(x)\mathrm{\mid }>2^{-k}\}$ is null, and $\mathrm{\mid }{f}_{k+1}-f_k\mathrm{\mid }\le 2^{-k}\text{on }X\setminus E_k\mathrm{.}$ With $E:={\bigcup }_k{E}_k$, [L6] makes $E$ measurable and null, and for $x\in X\setminus E$ and $m>n$, $\mathrm{\mid }{f}_m(x)-f_n(x)\mathrm{\mid }\le {\sum }_{k=n}^{m-1}{2}^{-k}\mathrm{.}$ So $(f_k(x))$ is Cauchy in $\mathbb{R}$, hence converges to some value $f(x)$. Set $f=0$ on $E$. The measurable functions equal to $f_k$ on $X\setminus E$ and zero on $E$ converge everywhere to $f$, so [L4] makes $f$ measurable. Indeed, their inverse images of Borel sets are obtained by intersecting those of $f_k$ with $X\setminus E$ and adjoining $E$ exactly when the Borel set contains zero. No completeness of the measure is assumed.

In the finite-$p$ case, $E:=\{g=\mathrm{\infty }\}$ is measurable and null: for every $M>0$, $M^p\mu (E)\le \int g^p<\mathrm{\infty }$, hence $\mu (E)=0$. On $X\setminus E$ the series ${\sum }_k\mathrm{\mid }{f}_{k+1}-f_k\mathrm{\mid }$ converges, so $f_k$ has a finite limit $f$. Set $f=0$ on $E$. Replacing each $f_k$ by zero on $E$ gives measurable functions converging everywhere to $f$, so [L4] gives measurability. Almost everywhere, $\mathrm{\mid }f\mathrm{\mid }\le \mathrm{\mid }{f}_0\mathrm{\mid }+g$, so [L2] gives $f\in {\mathcal{L}}^p$, and [step 1.1, L2, L3, L4] $\mathrm{\mid }f-f_k\mathrm{\mid }\le {\sum }_{j\ge k}{h}_j\le g\mathrm{.}$ Dominated convergence in [L3] therefore gives $\mathrm{\parallel }f-f_k{\mathrm{\parallel }}_p\to 0$. So the subsequence $u_{n_k}$ converges to $[f]$, and its representatives converge to $f$ almost everywhere.

For $x\in X\setminus E$, the same tail estimate gives [step 1.2] $\mathrm{\mid }f(x)-f_n(x)\mathrm{\mid }\le {\sum }_{k=n}^{\mathrm{\infty }}{2}^{-k}\le 2^{1-n}\mathrm{.}$ Taking $n=0$ also gives $\mathrm{\mid }f\mathrm{\mid }\le \mathrm{\mid }{f}_0\mathrm{\mid }+2$ almost everywhere, so $f\in {\mathcal{L}}^{\mathrm{\infty }}$. Therefore $\mathrm{\parallel }{u}_{n_k}-[f]{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. Given $\epsilon >0$, choose $K$ with $\mathrm{\parallel }{u}_n-u_m{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon \mathrm{/}2$ for $m,n\ge K$, then choose $k$ with $n_k\ge K$ and $\mathrm{\parallel }{u}_{n_k}-[f]{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon \mathrm{/}2$. Hence $\mathrm{\parallel }{u}_n-[f]{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }{u}_n-u_{n_k}{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }{u}_{n_k}-[f]{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon ,$ so $u_n\to [f]$ in $L^{\mathrm{\infty }}(\mu )$.

Since $(u_n)$ is Cauchy, for every $\epsilon >0$ there is $K$ with $\mathrm{\parallel }{u}_n-u_m{\mathrm{\parallel }}_p<\epsilon \mathrm{/}2$ whenever $m,n\ge K$. Choose $k$ with $n_k\ge K$ and $\mathrm{\parallel }{u}_{n_k}-[f]{\mathrm{\parallel }}_p<\epsilon \mathrm{/}2$ from step 2.1. Then for every $n\ge K$, [step 2.1, L2] $\mathrm{\parallel }{u}_n-[f]{\mathrm{\parallel }}_p\le \mathrm{\parallel }{u}_n-u_{n_k}{\mathrm{\parallel }}_p+\mathrm{\parallel }{u}_{n_k}-[f]{\mathrm{\parallel }}_p<\epsilon \mathrm{.}$ So the whole sequence converges to $[f]$. This proves completeness for $1\le p<\mathrm{\infty }$.

Step 2.1 proves the almost-everywhere convergent subsequence clause in the finite-$p$ case, and step 2.2 gives the same for $p=\mathrm{\infty }$. Steps 3.1 and 2.2 prove completeness in every case, which by [L1] is exactly completeness of the norm metric.