Integrating the John-Nirenberg tail recovers the Lq oscillation bound

Writing $A(\lambda ):=\mathrm{\mid }Q{\mathrm{\mid }}^{-1}\mathrm{\mid }\{x\in Q:\mathrm{\mid }b-b_Q\mathrm{\mid }>\lambda \}\mathrm{\mid }$, [L1] divided by $\mathrm{\mid }Q\mathrm{\mid }$ gives the identity $\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q\mathrm{\mid }b-b_Q{\mathrm{\mid }}^q=q{\int }_0^{\mathrm{\infty }}{\lambda }^{q-1}A(\lambda )d\lambda$.

If $\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}>0$, [L2] bounds $A(\lambda )\le C_n{e}^{-c_n\lambda \mathrm{/}\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}}$ for every $\lambda >0$, so the integral of step 1.1 is at most $q{C}_n{\int }_0^{\mathrm{\infty }}{\lambda }^{q-1}{e}^{-c_n\lambda \mathrm{/}\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}}d\lambda =q{C}_n\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}^q{\int }_0^{\mathrm{\infty }}{\mu }^{q-1}{e}^{-c_n\mu }d\mu$ after the substitution $\lambda =\mu \mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}$; the last integral is finite, so with $C_{n,q}:=q{C}_n{\int }_0^{\mathrm{\infty }}{\mu }^{q-1}{e}^{-c_n\mu }d\mu$ the asserted bound follows.

If $\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}=0$, then $b$ is constant almost everywhere by [L2], so $A(\lambda )=0$ for every $\lambda >0$ and the integral of step 1.1 vanishes; in particular the bound of step 2.1 holds with both sides $0$ for any finite $C_{n,q}$.

Steps 2.1 and 3.1 give $\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\int }_Q\mathrm{\mid }b-b_Q{\mathrm{\mid }}^q\le C_{n,q}\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{B}\mathrm{M}\mathrm{O}}^q$ for every cube, which is exactly the per-cube upper bound used in [L3]: taking $q$-th roots and the supremum over $Q$ yields the equivalence of the $L^q$ oscillation seminorms with the BMO seminorm.