shall now proceed to prove the stated results.

The functional $\sigma $ is obviously rearrangement invariant and, thanks to the Monotone Convergence Theorem, it satisfies the lattice axiom and the Fatou axiom. From (P1), only the triangle inequality needs proving. Let $f,g\in \mathcal{M}(\mathbb{R}^{n})$. By the definition of the associate space, one has $$\begin{aligned}
\left \|  \int _{t}^{\infty }(f+g)^{*}(s)\frac{{\mathrm{d}}s}{s}\right \|  _{X'(0,\infty )}
=
\sup _{\|h\|_{X(0,\infty )}\leq 1}
\int _{0}^{\infty }h(t)\int _{t}^{\infty }(f+g)^{*}(s)\frac{{\mathrm{d}}s}{s}\,{\mathrm{d}}t.\end{aligned}$$ Since the function $$\begin{aligned}
t\mapsto \int _{t}^{\infty }(f+g)^{*}(s)\frac{{\mathrm{d}}s}{s}\end{aligned}$$ is nonincreasing on $(0,\infty )$, we in fact have (cf. ) $$\begin{aligned}
\left \|  \int _{t}^{\infty }(f+g)^{*}(s)\frac{{\mathrm{d}}s}{s}\right \|  _{X'(0,\infty )}
=
\sup _{\|h\|_{X(0,\infty )}\leq 1}
\int _{0}^{\infty }h^{*}(t)\int _{t}^{\infty }(f+g)^{*}(s)\frac{{\mathrm{d}}s}{s}\,{\mathrm{d}}t.\end{aligned}$$ Thus, by the Fubini theorem, $$\begin{aligned}
\left \|  \int _{t}^{\infty }(f+g)^{*}(s)\frac{{\mathrm{d}}s}{s}\right \|  _{X'(0,\infty )}
&=
\sup _{\|h\|_{X(0,\infty )}
\leq 1}\int _{0}^{\infty }(f+g)^{*}(s)h^{**}(s)\,{\mathrm{d}}s.\end{aligned}$$ By  and the Hardy lemma, one has, for every such $h$, $$\begin{aligned}
\int _{0}^{\infty }(f+g)^{*}(s)h^{**}(s)\,{\mathrm{d}}s\leq \int _{0}^{\infty }f^{*}(s)h^{**}(s)\,{\mathrm{d}}s+\int _{0}^{\infty }g^{*}(s)h^{**}(s)\,{\mathrm{d}}s.\end{aligned}$$ This estimate, combined with the preceding identity and the subadditivity of the supremum, finally yields $$\begin{aligned}
\left \|  \int _{t}^{\infty }(f+g)^{*}(s)\frac{{\mathrm{d}}s}{s}\right \|  _{X'(0,{\infty })}
\leq \left \|  \int _{t}^{\infty
}f^{*}(s)\frac{{\mathrm{d}}s}{s}\right \|  _{X'(0,{\infty })}
+
\left \|  \int _{t}^{\infty }g
^{*}(s)\frac{{\mathrm{d}}s}{s}\right \|  _{X'(0,{\infty })},\end{aligned}$$ establishing the triangle inequality for $\sigma 