Well-definedness of the Slobodeckij seminorm and norm

Representative independence (i). Let $g=h$ almost everywhere and let $N:=\{x:g(x)\mathrm{\ne }h(x)\}$, a Lebesgue-null set. The difference $\mathrm{\mid }g(x)-g(y){\mathrm{\mid }}^p-\mathrm{\mid }h(x)-h(y){\mathrm{\mid }}^p$ vanishes whenever $x\mathrm{\notin }N$ and $y\mathrm{\notin }N$, so the two integrands differ at most on $E:=(N\times {\mathbb{R}}^d)\cup ({\mathbb{R}}^d\times N)$. Tonelli [F2] applied to the indicator of $N\times {\mathbb{R}}^d$ gives $({\lambda }_d\times {\lambda }_d)(N\times {\mathbb{R}}^d)={\int }_{{\mathbb{R}}^d}{\lambda }_d({\mathbb{R}}^d){\mathbf{1}}_N(x)dx=0$, because ${\mathbf{1}}_N$ vanishes off the null set $N$; the second piece is handled the same way, and ${\lambda }_d$ is sigma-finite. Hence $E$ is null and the two integrals coincide, possibly both infinite, so $[g{]}_{s,p}=[h{]}_{s,p}$.

Homogeneity (ii). For $\lambda \in \mathbb{K}$ one has $\mathrm{\mid }(\lambda g)(x)-(\lambda g)(y){\mathrm{\mid }}^p=\mathrm{\mid }\lambda {\mathrm{\mid }}^p\mathrm{\mid }g(x)-g(y){\mathrm{\mid }}^p$ pointwise, so the integrals are related by the factor $\mathrm{\mid }\lambda {\mathrm{\mid }}^p$ and $[\lambda g{]}_{s,p}=\mathrm{\mid }\lambda \mathrm{\mid }[g{]}_{s,p}$; at $\lambda =0$ both sides are $0$ while for $[g{]}_{s,p}=+\mathrm{\infty }$ and $\lambda \mathrm{\ne }0$ both sides are $+\mathrm{\infty }$.

Triangle inequality (ii). Put ${\mathrm{\Delta }}_g(x,y):=g(x)-g(y)$. Then $[{\mathrm{\Delta }}_{f+g}](x,y)={\mathrm{\Delta }}_f(x,y)+{\mathrm{\Delta }}_g(x,y)$ pointwise, and $[\cdot {]}_{s,p}$ is the $L^p$ norm of ${\mathrm{\Delta }}_{\cdot }$ on the sigma-finite measure space $({\mathbb{R}}^d\times {\mathbb{R}}^d,\mu )$ with $d\mu =\mathrm{\mid }x-y{\mathrm{\mid }}^{-d-sp}dxdy$. If $[f{]}_{s,p}+[g{]}_{s,p}=+\mathrm{\infty }$ the claim is trivial; otherwise Minkowski's integral inequality [F4] applied with $Y=\{1,2\}$ carrying counting measure and $F((x,y),1)={\mathrm{\Delta }}_f(x,y)$, $F((x,y),2)={\mathrm{\Delta }}_g(x,y)$ gives $[f+g{]}_{s,p}\le [f{]}_{s,p}+[g{]}_{s,p}$.

Zero seminorm forces almost-everywhere constancy (iii). Assume $[g{]}_{s,p}=0$, so by [F1] the nonnegative integrand $u(x,y):=\mathrm{\mid }g(x)-g(y){\mathrm{\mid }}^p\mathrm{\mid }x-y{\mathrm{\mid }}^{-d-sp}$ has integral $0$; by [F3] $u=0$ almost everywhere for the completed product measure, and since $\mathrm{\mid }x-y{\mathrm{\mid }}^{-d-sp}>0$ off the diagonal, $\mathrm{\mid }g(x)-g(y)\mathrm{\mid }=0$ for almost every pair $(x,y)$ in the product measure. Applying the Fubini clause of [F2] to the indicator of $E:=\{(x,y):g(x)\mathrm{\ne }g(y)\}$, whose product integral is $0$, gives a Lebesgue-null set $M$ such that $E_x$ is null for every $x\mathrm{\notin }M$. Fix $x_0\mathrm{\notin }M$ with $g(x_0)$ finite, possible because $g\in L^p$ is finite almost everywhere; then $g(y)=g(x_0)$ for almost every $y$.

Conclusion (iii) and the norm. By step 1.4 there is $c\in \mathbb{K}$ with $g=c$ almost everywhere. If $c\mathrm{\ne }0$, then ${\int }_{{\mathbb{R}}^d}\mathrm{\mid }g{\mathrm{\mid }}^p=\mathrm{\mid }c{\mathrm{\mid }}^p{\lambda }_d({\mathbb{R}}^d)$, and ${\lambda }_d({\mathbb{R}}^d)=+\mathrm{\infty }$ because ${\lambda }_d({\mathbb{R}}^d)\ge {\lambda }_d([0,R{]}^d)=R^d$ for every $R>0$ by [F5] and monotonicity of a measure; this contradicts $g\in L^p$. Hence $c=0$, and $[g{]}_{s,p}=0$ forces $g=0$ almost everywhere. Consequently $\mathrm{\parallel }g{\mathrm{\parallel }}_{W^{s,p}}=\mathrm{\parallel }g{\mathrm{\parallel }}_{L^p}+[g{]}_{s,p}$ vanishes only on the zero class, is homogeneous by step 1.2 and the homogeneity of the $L^p$ norm [F6, F7], and satisfies the triangle inequality by step 1.3, the triangle inequality of the $L^p$ norm [F6, F7], and addition of inequalities; $W^{s,p}({\mathbb{R}}^d)$ is a vector space because sums and scalar multiples of classes with finite $L^p$ norm and finite seminorm again have $L^p$ norm and seminorm finite by steps 1.2 and 1.3. For the analogue over a finite-measure set $\mathrm{\Omega }$ the integrand of a constant is identically zero, so the seminorm alone vanishes on constants and only the sum norm is definite; adding the $L^p$ term removes that ambiguity, as claimed.