(5), p. 18 — g_T(ω_T) ≥ g_T(ω_S) gives b_S − b_T ≤ a_T − a_S

Let $\omega_S=(p_S,q_S,\bar v_S)$ and $\omega_T=(p_T,q_T,\bar v_T)$ be points of $\Omega$, and write $a_i=\frac{1}{1-p_i}$, $b_i=\frac{1}{1-q_i}$ for $i\in\{S,T\}$. If

that is $g_T(\omega_T)\ge g_T(\omega_S)$, then

In the proof of Theorem 4.2 this is applied to the engagement maximizer $\omega_T$ and the utility maximizer $\omega_S$, which is feasible for (4).

Formalization Note The paper prints (5) with a strict inequality, from "$\mathbb E[T(\omega_T)]>\mathbb E[T(\omega_S)]$". The definition (4) of $\omega_T$ gives only $\ge$, so (5) is posed with $\le$ under the hypothesis $g_T(\omega_T)\ge g_T(\omega_S)$; the conclusion of Theorem 4.2 (a non-strict inequality) is unaffected.