<1$ there exists $C>0$ such that $\psi(t)\le C t^\varepsilon$ for all $t>0$. Hence, $$c_H  \ge \frac{a}{t C a^\varepsilon}=\frac{a^{1-\varepsilon}}{C t}.$$

Therefore, $$\mu_H(1/t) \le (C c_H t)^\frac{1}{1-\varepsilon}.$$

Since this inequality holds for every $0<\varepsilon <1$, it follows that for every $\alpha>1$ and sufficiently large $t>0$ one has $\mu_H(1/t) \le t^\alpha.$

\(b) For sufficiently large $t>0$ one has $$\int_0^t H(s) ds \le \int_0^{\mu_H(1/t)} H(s) ds +1 \le \int_0^{t^\alpha} H(s) ds+1,$$ where the first inequality was proved in [@EZ Proposition 8] and the second one was proved above.

Dividing by $\psi(t)$ and applying extended limits, yields $$\label{eq2}
\lim_{t\to \omega} \frac1{\psi(t)} \int_0^t H(s) ds\le\lim_{t\to \omega} \frac1{\psi(t)} \int_0^{\mu_H(1/t)} H(s) ds \le\lim_{t\to \omega} \frac1{\psi(t)} \int_0^{t^\alpha} H(s) ds.$$

Since $\psi$ satisfies , it follows from the property of extended limits that $$\begin{aligned}
\lim_{t\to \omega} \frac1{\psi(t^\alpha)} \int_0^{t^\alpha} H(s) ds&\le \lim_{t\to \omega} \frac{\psi(t^\alpha)}{\psi(t)} \frac1{\psi(t^\alpha)} \int_0^{t^\alpha} H(s) ds\\ 
&=A_\psi(\alpha) \lim_{t\to \omega}  \frac1{\psi(t^\alpha)} \int_0^{t^\alpha} H(s) ds.\end{aligned}$$

Since $\omega$ is exponentiation invariant, it follows that $$\label{eq3}
\lim_{t\to \omega} \frac1{\psi(t^\alpha)} \int_0^{t^\alpha} H(s) ds\le A_\psi(\alpha) \lim_{t\to \omega}  \frac1{\psi(t)} \int_0^{t} H(s) ds.$$

Combining  and , we obtain that $$\lim_{t\to \omega} \frac1{\psi(t)} \int_0^t H(s) ds\le\lim_{t\to \omega} \frac1{\psi(t)} \int_0^{\mu_H(1/t)} H(s) ds \le A_\psi(\alpha) \lim_{t\to \omega}  \frac1{\psi(t)} \int_0^{t} H(s) ds$$ holds for every $\alpha>1$. It follows from [@GS Lemma 