The $L^p$ distance for $0<p<1$ is a complete translation-invariant metric

If $f\sim f^{\mathrm{\prime }}$ and $g\sim g^{\mathrm{\prime }}$, then $f-f^{\mathrm{\prime }}$ and $g-g^{\mathrm{\prime }}$ vanish almost everywhere. Outside the union of those two null sets, one has $f-g=f^{\mathrm{\prime }}-g^{\mathrm{\prime }}$, so $\mathrm{\mid }f-g{\mathrm{\mid }}^p=\mathrm{\mid }{f}^{\mathrm{\prime }}-g^{\mathrm{\prime }}{\mathrm{\mid }}^p$ almost everywhere. Hence $d_p([f],[g])$ is well defined. The same union-of-null-sets argument shows that addition and scalar multiplication descend to the quotient classes, and the inequality in [L7] shows that ${\mathcal{L}}^p(\mu )$ is closed under those operations.

Symmetry of $d_p$ is immediate. If $d_p([f],[g])=0$, then $\int \mathrm{\mid }f-g{\mathrm{\mid }}^{p}d\mu =0$, so $f=g$ almost everywhere and hence $[f]=[g]$. For the triangle inequality, the pointwise inequality from [L7] gives [L2, L4, L7] $\mathrm{\mid }f-h{\mathrm{\mid }}^p=\mathrm{\mid }(f-g)+(g-h){\mathrm{\mid }}^p\le \mathrm{\mid }f-g{\mathrm{\mid }}^p+\mathrm{\mid }g-h{\mathrm{\mid }}^p,$ and integrating yields $d_p([f],[h])\le d_p([f],[g])+d_p([g],[h])\mathrm{.}$ Thus [L2] makes $d_p$ a metric.

Let $(u_n)$ be Cauchy in $d_p$. Choose by least indices a subsequence $(u_{n_k})$ with [L3, L5, given, choose] $d_p(u_{n_{k+1}},u_{n_k})<2^{-k}\mathrm{.}$ By [A1], choose a sequence of representatives $f_k$ of $u_{n_k}$ and define $h_k:=\mathrm{\mid }{f}_{k+1}-f_k{\mathrm{\mid }}^p,g_m:={\sum }_{j<m}{h}_j\mathrm{.}$ Each $h_k$ is measurable and integrable, and [L3] gives a measurable pointwise limit $g:={\sum }_{j=0}^{\mathrm{\infty }}{h}_j$ with $\int gd\mu ={\lim }_{m\to \mathrm{\infty }}\int g_{m}d\mu \le {\sum }_{j=0}^{\mathrm{\infty }}{2}^{-j}<\mathrm{\infty }\mathrm{.}$ Hence $g<\mathrm{\infty }$ almost everywhere. This is the only use of Choice in the argument.

Translation invariance is pointwise: [step 1.1] $d_p([f]+[u],[g]+[u])=\int \mathrm{\mid }(f+u)-(g+u){\mathrm{\mid }}^{p}d\mu =d_p([f],[g])\mathrm{.}$

Put $E:=\{x:g(x)=\mathrm{\infty }\}$, a measurable null set by step 1.3. Fix $x\mathrm{\notin }E$. Then ${\sum }_j\mathrm{\mid }{f}_{j+1}(x)-f_j(x){\mathrm{\mid }}^p<\mathrm{\infty }$, so the terms tend to $0$. Thus $\mathrm{\mid }{f}_{j+1}(x)-f_j(x)\mathrm{\mid }\le 1$ for all large $j$, and then [step 1.3, L3, L7] $\mathrm{\mid }{f}_{j+1}(x)-f_j(x)\mathrm{\mid }\le \mathrm{\mid }{f}_{j+1}(x)-f_j(x){\mathrm{\mid }}^p=h_j(x)\mathrm{.}$ So the real series ${\sum }_j\mathrm{\mid }{f}_{j+1}(x)-f_j(x)\mathrm{\mid }$ converges by comparison with ${\sum }_j{h}_j(x)$, which makes $(f_k(x))$ converge to some real value $f(x)$. Define $f=0$ on $E$. The measurable functions equal to $f_k$ on $X\setminus E$ and zero on $E$ converge pointwise to $f$, so [L3] and [L5] make $f$ measurable. Moreover, $\mathrm{\mid }f{\mathrm{\mid }}^p\le \mathrm{\mid }{f}_0{\mathrm{\mid }}^p+g$ on $X\setminus E$ by [L7], and $f=0$ on $E$. Since $f_0\in {\mathcal{L}}^p(\mu )$ and $g$ is integrable, $f\in {\mathcal{L}}^p(\mu )$, so $[f]$ is a valid element of $L^p(\mu )$. Also $\mathrm{\mid }f(x)-f_k(x){\mathrm{\mid }}^p\le {({\sum }_{j\ge k}\mathrm{\mid }{f}_{j+1}(x)-f_j(x)\mathrm{\mid })}^p\le {\sum }_{j\ge k}{h}_j(x)\mathrm{.}$ Integrating and using monotone convergence on the tails yields $d_p(u_{n_k},[f])\le {\sum }_{j\ge k}{d}_p(u_{n_{j+1}},u_{n_j})\le {\sum }_{j\ge k}{2}^{-j}\to 0.$

Because $(u_n)$ is Cauchy, given $\epsilon >0$ choose $K$ with $d_p(u_n,u_m)<\epsilon \mathrm{/}2$ for $m,n\ge K$, then choose $k$ with $n_k\ge K$ and $d_p(u_{n_k},[f])<\epsilon \mathrm{/}2$ from step 2.2. The triangle inequality from step 1.2 gives $d_p(u_n,[f])<\epsilon$ for all $n\ge K$. Hence $(L^p(\mu ),d_p)$ is complete.

Steps 1.2 and 2.1 prove that $d_p$ is a translation-invariant metric, and step 3.1 proves completeness.