 Solution

Write $A_q=⟨f{⟩}_{w,q}$ and $N_q=\Vert f{\Vert }_{w,q}$. The exponent on the finite-set expression is $-1/q^{\prime }$ in the PDF; the TeX aid drops its prime. Ignore zero-measure sets, whose integrals vanish. The equivalence isFor the first bound, let $E_a=\{|f|>a\}$ and choose a finite-measure subset $E\subset E_a$. Since ${\int }_E|f|\ge a|E|$,Exhaust $E_a$ by its intersections with bounded balls. If its measure is finite these give the desired bound $a|E_a{|}^{1/q}$ in the limit; if it is infinite they force $N_q=\infty$. Now take the supremum over $a$.

For the other direction it suffices to assume $0<A_q<\infty$; the zero and infinite cases are immediate. For a set $E$ with $0<|E|=m<\infty$, the layer cake representation givesSplit at $t_0=A_q{m}^{-1/q}$. The first contribution is $m{t}_0$, and the second is $A_q^q{t}_0^{1-q}/(q-1)$. Their sum is $q^{\prime }{A}_q{m}^{1/q^{\prime }}$. Multiply by $m^{-1/q^{\prime }}$ and take the supremum. This proves the equivalence even when the quantities are extended-valued. The integral expression is an actual norm on the weak Lq space, since the triangle inequality holds inside every set integral. The distribution-function expression is generally a quasi-norm. The factor $q^{\prime }$ is sharp: $f(x)=x^{-1/q}$ on $(0,\infty )$ has $A_q=1$ and attains $N_q=q^{\prime }$ on every initial interval. In higher dimensions, multiply this example by the indicator of a unit box in the remaining coordinates.