A.1. Proof of Theorem 1
With a simple algebra we obtain
$$\log\bigl{\{}\mathbf{E}\exp\bigl{[}{\textrm{i}}tB_{h}(\xi)\bigr{]}\bigr{\}}=n_{h}\log\left\{\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty}\cos\biggl{[}\frac{tS(x)}{n_{h}}\biggr{]}{\textrm{e}}^{-x^{2}/2}\ dx+\frac{\textrm{i}}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty}\sin\biggl{[}\frac{tS(x)}{n_{h}}\biggr{]}{\textrm{e}}^{-{x^{2}}/{2}}\ dx\right\}$$
$${}=n_{h}\log\left\{1+\int\limits_{-\infty}^{\infty}\biggr{[}\cos\biggl{[}\frac{t}{n_{h}}\biggl{(}\frac{1}{\Phi(x)}-1\biggr{)}\biggr{]}-1\biggl{]}\ d\Phi(x)+{\textrm{i}}\int\limits_{-\infty}^{\infty}\sin\biggl{[}\frac{t}{n_{h}}\biggl{(}\frac{1}{\Phi(x)}-1\biggr{)}\biggr{]}\ d\Phi(x)\right\}$$
$${}=n_{h}\log\left\{1+\int\limits_{0}^{1}\biggl{\{}\cos\biggl{[}\frac{t}{n_{h}}\biggl{(}\frac{1}{u}-1\biggr{)}\biggr{]}-1\biggr{\}}\ du+{\textrm{i}}\int\limits_{0}^{1}\sin\biggl{[}\frac{t}{n_{h}}\biggl{(}\frac{1}{u}-1\biggr{)}\biggr{]}\ du\right\}$$
$${}=n_{h}\log\left\{1+\int\limits_{0}^{\infty}\frac{1}{(1+u)^{2}}\biggl{[}\cos\biggl{(}\frac{tu}{n_{h}}\biggr{)}-1\biggr{]}\ du+{\textrm{i}}\int\limits_{0}^{\infty}\frac{1}{(1+u)^{2}}\sin\biggl{(}\frac{tu}{n_{h}}\biggr{)}\ du\right\}$$
$${}=n_{h}\log\left\{1+\frac{|t|}{n_{h}}\int\limits_{0}^{\infty}\frac{\cos(u)-1}{(|t|/n_{h}+u)^{2}}\ du+\frac{{\textrm{i}}t}{n_{h}}\int\limits_{0}^{\infty}\frac{\sin(u)}{(|t|/n_{h}+u)^{2}}\ du\right\}.$$
(A.1)
It is clear that as \(n_{h}\rightarrow\infty\)
$$\int\limits_{0}^{\infty}\frac{\cos(u)-1}{(|t|/n_{h}+u)^{2}}\ du\rightarrow\int\limits_{0}^{\infty}\frac{\cos(u)-1}{u^{2}}\ du=-\frac{\pi}{2}.$$
(A.2)
Let us choose \(\epsilon_{h}<1\) such that
$$\lim_{h\rightarrow 0}\epsilon_{h}=0,\quad\lim_{h\rightarrow 0}\epsilon_{h}n_{h}=\infty.$$
Then we get by the Taylor formula
$$\int\limits_{0}^{\infty}\frac{\sin(u)}{(|t|/n_{h}+u)^{2}}\ du=\int\limits_{0}^{\epsilon_{h}}\frac{\sin(u)}{(|t|/n_{h}+u)^{2}}\ du+\int\limits_{\epsilon_{h}}^{\infty}\frac{\sin(u)}{(|t|/n_{h}+u)^{2}}\ du$$
$${}=\int\limits_{0}^{\epsilon_{h}}\frac{u}{(|t|/n_{h}+u)^{2}}\ du+O\left[\int\limits_{0}^{\epsilon_{h}}\frac{u^{3}}{(|t|/n_{h}+u)^{2}}\ du\right]+(1+o(1))\int\limits_{\epsilon_{h}}^{\infty}\frac{\sin(u)}{u^{2}}\ du$$
$${}=\log\bigg{(}1+\frac{\epsilon_{h}n_{h}}{|t|}\biggr{)}-\frac{\epsilon_{h}n_{h}}{(|t|+\epsilon_{h}n_{h})}+(1+o(1))\int\limits_{\epsilon_{h}}^{\infty}\frac{\sin(u)}{u^{2}}\ du+O(\epsilon_{h}^{2}).$$
(A.3)
Next, integrating by parts, we obtain
$$\int\limits_{x}^{\infty}\frac{\sin(z)}{z^{2}}\ dz=-\int\limits_{x}^{\infty}\frac{\sin(z)}{z}\ d\biggl{(}\log\frac{1}{z}\biggr{)}=\frac{\sin(x)}{x}\log\frac{1}{x}+\int\limits_{x}^{\infty}\frac{z\cos(z)-\sin(z)}{z^{2}}\log\frac{1}{z}\ dz.$$
Hence, as \(x\rightarrow 0\)
$$\int\limits_{x}^{\infty}\frac{\sin(z)}{z^{2}}\ dz=\log\frac{1}{x}+\int\limits_{0}^{\infty}\frac{z\cos(z)-\sin(z)}{z^{2}}\log\frac{1}{z}\ dz+O\bigg{(}x^{2}\log\frac{1}{x}\biggr{)}$$
$${}=\log\frac{1}{x}+(1-\gamma)+O\bigg{(}x^{2}\log\frac{1}{x}\biggr{)},$$
where \(\gamma\) is Euler’s constant.
With this equation we continue (44) as follows:
$$\int\limits_{0}^{\infty}\frac{\sin(u)}{(|t|/n_{h}+u)^{2}}\ du=\log\frac{1}{|t|}+\log(n_{h})-\gamma+O\biggl{(}\frac{|t|}{\epsilon_{h}n_{h}}\biggr{)}+O\biggl{(}\epsilon_{h}^{2}\log\frac{1}{\epsilon_{h}}\biggr{)}.$$
Substituting this equation and (43) in (42), we get
$$\log\bigl{\{}\mathbf{E}\exp\bigl{[}{\textrm{i}}tB_{h}(\xi)\bigr{]}\bigr{\}}=-\frac{\pi|t|}{2}+{\textrm{i}}t\biggl{(}\log\frac{1}{|t|}+\log(n_{h})-\gamma\biggr{)}+o(1),$$
thus, proving the theorem.
A.2. Proof of Theorem 3
I. A lower bound. By (22) we have for any given \(x\)
$$\beta^{M}_{\rho,\tau}(A)\geqslant\mathbf{P}\biggl{\{}\biggl{[}\log\biggl{[}S\biggl{(}-\frac{A}{\sigma_{\rho}}+\xi_{\rho,\tau}\biggr{)}\biggr{]}-\log\frac{n_{\rho}}{\bar{\pi}_{\rho}}\biggr{]}$$
$${}\bigvee\biggl{[}\zeta_{\rho}-\log\frac{1}{\bar{\pi}_{\rho}}\biggr{]}\leqslant q_{\alpha}^{\varkappa}\biggr{\}}\mathbf{P}\biggl{\{}\max_{h\in\mathcal{H}}\biggl{[}\zeta_{h}-\log\frac{1}{\bar{\pi}_{h}}\biggr{]}\leqslant q_{\alpha}^{\varkappa}\biggr{\}}$$
$${}=\mathbf{P}\biggl{\{}S\biggl{(}-\frac{A}{\sigma_{\rho}}+\xi_{\rho,\tau}\biggr{)}\leqslant\exp\biggl{[}q_{\alpha}^{\varkappa}+\log\frac{n_{\rho}}{\bar{\pi}_{\rho}}\biggr{]}\biggr{\}}\mathbf{P}\biggl{\{}\zeta_{\rho}\leqslant q_{\alpha}^{\varkappa}+\log\frac{1}{\bar{\pi}_{\rho}}\biggr{\}}\mathbf{P}\bigl{\{}\zeta\leqslant q_{\alpha}^{\varkappa}\bigr{\}}$$
$${}\geqslant(1-\alpha)^{1+\bar{\pi}_{\rho}}\mathbf{P}\biggl{\{}S\biggl{(}-\frac{A}{\sigma_{\rho}}+\xi_{\rho,\tau}\biggr{)}\leqslant\exp\biggl{[}q_{\alpha}^{\varkappa}+\log\frac{n_{\rho}}{\bar{\pi}_{\rho}}\biggr{]};\xi_{\rho,\tau}\geqslant x\biggr{\}}$$
$${}\geqslant(1-\alpha)^{1+\bar{\pi}_{\rho}}\mathbf{P}\bigl{\{}\xi_{\rho,\tau}\geqslant x\bigr{\}}\mathbf{1}\biggl{\{}S\biggl{(}-\frac{A}{\sigma_{\rho}}+x\biggr{)}\leqslant\exp\biggl{[}q_{\alpha}^{\varkappa}+\log\frac{n_{\rho}}{\bar{\pi}_{\rho}}\biggr{]}\biggr{\}}.$$
(A.4)
Let \(R(z)\geqslant 0\) be a solution to
$$S\Bigl{[}-\sqrt{R(z)}\Bigr{]}=z.$$
It is easy to check with the help of (7) that as
\(z\rightarrow\infty\)
$$R(z)=2\log(z)-\log[4\pi\log(z)]+o(1).$$
(A.5)
Denote for brevity
$$r_{h}(q,u)=2\biggl{(}q+\log\frac{n_{h}}{u}\biggr{)}-\log\biggl{[}4\pi\biggl{(}q+\log\frac{n_{h}}{u}\biggr{)}\biggr{]}.$$
With (A.5) and the Markov inequality we obtain for any
\(\epsilon>0\)
$$\mathbf{E}_{\bar{\pi}}\biggl{[}\mathbf{1}\biggl{\{}S\biggl{(}-\frac{A_{h}}{\sigma_{h}}+x\biggr{)}\leqslant\exp\biggl{[}q_{\alpha}^{\varkappa}+\log\frac{n_{h}}{\bar{\pi}_{h}}\biggr{]}\biggr{\}}\biggr{]}=1-\mathbf{E}_{\bar{\pi}}\biggl{[}\mathbf{1}\biggl{\{}S\biggl{(}-\frac{A_{h}}{\sigma_{h}}+x\biggr{)}>\exp\biggl{[}q_{\alpha}^{\varkappa}+\log\frac{n_{h}}{\bar{\pi}_{h}}\biggr{]}\biggr{\}}\biggr{]}$$
$${}=1-\mathbf{E}_{\bar{\pi}}\biggl{[}\mathbf{1}\biggl{\{}\biggl{(}\frac{A_{h}}{\sigma_{h}}-x\biggr{)}^{2}-r_{h}(q_{\alpha}^{\varkappa},\bar{\pi}_{h})>o(1)\biggr{\}}\biggr{]}$$
$${}\geqslant 1-\frac{1}{\epsilon H^{*}_{\bar{\pi}}}\mathbf{E}_{\bar{\pi}}\biggl{[}\biggl{(}\frac{A_{h}}{\sigma_{h}}-x\biggr{)}^{2}-r_{h}(q_{\alpha}^{\varkappa},\bar{\pi}_{h})+\epsilon H^{*}_{\bar{\pi}}\biggr{]}_{+}.$$
(A.6)
With condition (25) and simple algebra it is easy to check that
$$\lim_{\bar{\pi}\rightarrow 0}\frac{1}{H^{*}_{\bar{\pi}}}\mathbf{E}_{\bar{\pi}}\biggl{|}r_{h}(q_{\alpha}^{\varkappa},\bar{\pi})-R_{h}(q_{\alpha}^{\varkappa},H_{\bar{\pi}})\biggr{|}=0.$$
Therefore (29) follows from (45), (47) and the above equation.
II. An upper bound. Since \(S(x)\) is a
decreasing function, we have obviously for any \(x\geqslant 0\)
$$\bar{\beta}^{M}_{\bar{\pi}}(\mathbf{A})\leqslant\mathbf{E}_{\bar{\pi}}\biggl{[}\mathbf{P}\biggl{\{}S\biggl{(}-\frac{A_{h}}{\sigma_{h}}+\xi\biggr{)}\leqslant\exp\biggl{[}q_{\alpha}^{\varkappa}+\log\frac{n_{h}}{\bar{\pi}_{h}}\biggr{]}\biggr{\}}\biggr{]}$$
$${}\leqslant\mathbf{E}_{\bar{\pi}}\biggl{[}\mathbf{P}\biggl{\{}S\biggl{(}-\frac{A_{h}}{\sigma_{h}}+\xi\biggr{)}\leqslant\exp\biggl{[}q_{\alpha}^{\varkappa}+\log\frac{n_{h}}{\bar{\pi}_{h}}\biggr{]};\xi\leqslant x\biggr{\}}\biggr{]}$$
$${}+\mathbf{E}_{\bar{\pi}}\biggl{[}\mathbf{P}\biggl{\{}S\biggl{(}-\frac{A_{h}}{\sigma_{h}}+\xi\biggr{)}\leqslant\exp\biggl{[}q_{\alpha}^{\varkappa}+\log\frac{n_{h}}{\bar{\pi}_{h}}\biggr{]};\xi>x\biggr{\}}\biggr{]}$$
$${}\leqslant\mathbf{E}_{\bar{\pi}}\biggl{[}\mathbf{1}\biggl{\{}S\biggl{(}-\frac{A_{h}}{\sigma_{h}}+x\biggr{)}\leqslant\exp\biggl{[}q_{\alpha}^{\varkappa}+\log\frac{n_{h}}{\bar{\pi}_{h}}\biggr{]}\biggr{\}}\biggr{]}+\mathbf{P}\bigl{\{}\xi>x\bigr{\}}.$$
Next, with (46), the Markov inequality, and Condition (25) we get
for any \(\epsilon>0\)
$$\sum_{h\in\mathcal{H}}\bar{\pi}_{h}\mathbf{1}\biggl{\{}S\biggl{(}-\frac{A_{h}}{\sigma_{h}}+x\biggr{)}\leqslant\exp\biggl{[}q_{\alpha}^{\varkappa}+\log\frac{n_{h}}{\bar{\pi}_{h}}\biggr{]}\biggr{\}}$$
$${}\leqslant\sum_{h\in\mathcal{H}}\bar{\pi}_{h}\mathbf{1}\biggl{\{}r_{h}(q_{\alpha}^{\varkappa},\bar{\pi}_{h})-\biggl{(}\frac{A_{h}}{\sigma_{h}}-x\biggr{)}^{2}>o(1)\biggr{\}}$$
$${}\leqslant\frac{1}{\epsilon H^{*}_{\bar{\pi}}}\sum_{h\in\mathcal{H}}\bar{\pi}_{h}\biggr{[}r_{h}(q_{\alpha}^{\varkappa},\bar{\pi}_{h})-\biggl{(}\frac{A_{h}}{\sigma_{h}}-x\biggr{)}^{2}+\epsilon H^{*}_{\bar{\pi}}\biggr{]}_{+}$$
$${}\leqslant\frac{1}{\epsilon H^{*}_{\bar{\pi}}}\sum_{h\in\mathcal{H}}\bar{\pi}_{h}\biggr{[}R_{h}(q_{\alpha}^{\varkappa},H_{\bar{\pi}})-\biggl{(}\frac{A_{h}}{\sigma_{h}}-x\biggr{)}^{2}+\epsilon H^{*}_{\bar{\pi}}\biggr{]}_{+}+o(1).$$
To complete the proof, let us choose \(x=\sqrt{\epsilon H^{*}_{\bar{\pi}}}\).
A.3. Proof of Theorem 4
A lower bound. For given \(x\) and \(\delta>0\) by (17) and
(13), we obtain
$$\beta^{B}_{\rho,\tau}(A)\geqslant\mathbf{P}\biggl{\{}\frac{\bar{\pi}_{\rho}}{n_{\rho}}S\biggl{(}-\frac{A}{\sigma_{\rho}}+\xi_{\rho,\tau}\biggr{)}+\zeta^{\circ}\leqslant q_{\alpha}^{\circ};\xi_{\rho,\tau}\geqslant x\biggr{\}}$$
$${}\geqslant\mathbf{P}\biggl{\{}\frac{\bar{\pi}_{\rho}}{n_{\rho}}S\biggl{(}-\frac{A}{\sigma_{\rho}}+x\biggr{)}+\zeta^{\circ}\leqslant q_{\alpha}^{\circ};\xi_{\rho,\tau}\geqslant x\biggr{\}}$$
$${}=\mathbf{P}\bigl{\{}\xi_{\rho,\tau}\geqslant x\bigr{\}}\mathbf{P}\biggl{\{}\zeta^{\circ}\leqslant q_{\alpha}^{\circ}-\frac{\bar{\pi}_{\rho}}{n_{\rho}}S\biggl{(}-\frac{A}{\sigma_{\rho}}+x\biggr{)}\biggr{\}}$$
$${}\geqslant\mathbf{P}\bigl{\{}\xi_{\rho,\tau}\geqslant x\bigr{\}}\mathbf{P}\bigl{\{}\zeta^{\circ}\leqslant(1-\delta)q_{\alpha}^{\circ}\bigr{\}}\mathbf{1}\biggl{\{}\frac{\bar{\pi}_{\rho}}{n_{\rho}}S\biggl{(}-\frac{A}{\sigma_{\rho}}+x\biggr{)}\leqslant\delta q_{\alpha}^{\circ}\biggr{\}}.$$
Similarly to (47), since \(\bar{\pi}\rightarrow 0\), we get
$$\mathbf{E}_{\bar{\pi}}\biggl{[}\mathbf{1}\biggl{\{}\frac{\bar{\pi}_{h}}{n_{h}}S\biggl{(}-\frac{A_{h}}{\sigma_{h}}+x\biggr{)}\leqslant\delta q_{\alpha}^{\circ}\biggr{\}}\biggr{]}$$
$${}\geqslant 1-\frac{1}{\epsilon H^{*}_{\bar{\pi}}}\mathbf{E}_{\bar{\pi}}\biggl{[}\biggl{(}\frac{A_{h}}{\sigma_{h}}-x\biggr{)}^{2}+\epsilon H^{*}_{\bar{\pi}}-R_{h}[\log(\delta q_{\alpha}^{\circ}),H_{\bar{\pi}}]\biggr{]}_{+}$$
$${}\geqslant 1-\frac{1}{\epsilon H^{*}_{\bar{\pi}}}\mathbf{E}_{\bar{\pi}}\biggl{[}\biggl{(}\frac{A_{h}}{\sigma_{h}}-x\biggr{)}^{2}+\epsilon H^{*}_{\bar{\pi}}-R_{h}[\log(q_{\alpha}^{\circ}),H_{\bar{\pi}}]\biggr{]}_{+}v\geqslant 1+o(1).$$
So, since \(\delta\) is arbitrary, (33) follows from the above
inequalities.
An upper bound. Since \(S(x)\) is decreasing, we
get
$$\beta^{B}_{\rho,\tau}(A)\leqslant\mathbf{P}\biggl{\{}\frac{\bar{\pi}_{\rho}}{n_{\rho}}S\biggl{(}-\frac{A}{\sigma_{\rho}}+\xi_{\rho,\tau}\biggr{)}+\zeta^{\circ}\leqslant q_{\alpha}^{\circ};\xi_{\rho,\tau}\geqslant x\biggr{\}}$$
$${}+\mathbf{P}\biggl{\{}\frac{\bar{\pi}_{\rho}}{n_{\rho}}S\biggl{(}-\frac{A}{\sigma_{\rho}}+\xi_{\rho,\tau}\biggr{)}+\zeta^{\circ}\leqslant q_{\alpha}^{\circ};\xi_{\rho,\tau}\leqslant x\biggr{\}}$$
$${}\leqslant\mathbf{P}\bigl{\{}\xi_{\rho,\tau}\geqslant x\bigr{\}}+\mathbf{P}\biggl{\{}\frac{\bar{\pi}_{\rho}}{n_{\rho}}S\biggl{(}-\frac{A}{\sigma_{\rho}}+x\biggr{)}+\zeta^{\circ}\leqslant q_{\alpha}^{\circ}\biggr{\}}.$$
(A.7)
Next, for any given \(x^{\circ}<q_{\alpha}^{\circ}\) we obtain with the
help of the Markov inequality and (46)
$$\mathbf{E}_{\bar{\pi}}\biggl{[}\mathbf{P}\biggl{\{}\frac{\bar{\pi}_{h}}{n_{h}}S\biggl{(}-\frac{A_{h}}{\sigma_{h}}+x\biggr{)}+\zeta^{\circ}\leqslant q_{\alpha}^{\circ}\biggr{\}}\biggr{]}$$
$${}\leqslant\mathbf{E}_{\bar{\pi}}\biggl{[}\mathbf{P}\biggl{\{}\frac{\bar{\pi}_{h}}{n_{h}}S\biggl{(}-\frac{A_{h}}{\sigma_{h}}+x\biggr{)}+\zeta^{\circ}\leqslant q_{\alpha}^{\circ};\zeta^{\circ}<x^{\circ}\biggr{\}}\biggr{]}$$
$${}+\mathbf{E}_{\bar{\pi}}\biggl{[}\mathbf{P}\biggl{\{}\frac{\bar{\pi}_{h}}{n_{h}}S\biggl{(}-\frac{A_{h}}{\sigma_{h}}+x\biggr{)}+\zeta^{\circ}\leqslant q_{\alpha}^{\circ};\zeta^{\circ}\geqslant x^{\circ}\biggr{\}}\biggr{]}$$
$${}\leqslant\mathbf{P}\bigl{\{}\zeta^{\circ}<x^{\circ}\}+\mathbf{E}_{\bar{\pi}}\biggl{[}\mathbf{1}\biggl{\{}\frac{\bar{\pi}_{h}}{n_{h}}S\biggl{(}-\frac{A_{h}}{\sigma_{h}}+x\biggr{)}\leqslant q_{\alpha}^{\circ}-x^{\circ}\biggr{\}}\biggr{]}$$
$${}\leqslant\mathbf{P}\bigl{\{}\zeta^{\circ}<x^{\circ}\}+\frac{1}{\epsilon H^{*}_{\bar{\pi}}}\mathbf{E}_{\bar{\pi}}\biggl{[}R_{h}[\log(q_{\alpha}^{\circ}-x^{\circ}),H_{\bar{\pi}}]+\epsilon H^{*}_{\bar{\pi}}-\biggl{(}\frac{A_{h}}{\sigma_{h}}-x\biggr{)}^{2}\biggr{]}_{+}.$$
Finally choosing \(x=\sqrt{\epsilon H^{*}_{\bar{\pi}}}\) and combining
this equation with (48), we complete the proof of (35).
A.4. Proof of Theorem 5
Let us consider
$$H\bigl{(}\bar{\pi}^{\omega,\nu},{\Pi}\bigr{)}=\sum_{h\in\mathcal{H}}\bar{\pi}^{\omega,\nu}_{h}\log\frac{1}{{\Pi}_{h}}.$$
One can check easily with (38)–(40) and (26) that as
\(\omega\rightarrow\infty\)
$$H\bigl{(}\bar{\pi}^{\omega,\nu},{\Pi}\bigr{)}=\log\frac{1}{\varepsilon}+\frac{1}{\omega}\sum_{k=1}^{\infty}\nu\biggl{(}\frac{k}{\omega}\biggr{)}L^{m,\varepsilon}(k)+o(1)=\log\frac{1}{\varepsilon}+o(1)$$
$${}+\int\limits_{0}^{\infty}\nu(x)\biggl{[}\log(x\omega+1)+\sum_{s=0}^{m-1}\log[\psi_{s}(x\omega)]+(1+\varepsilon)\log[\psi_{m}(x\omega)]\biggr{]}\ dx.$$
It is also clear in view of (27) that
$$\int\limits_{0}^{\infty}\nu(x)\log(x\omega+1)\ dx=\log(\omega)+O(1)$$
and for any integer \(s\geqslant 0\)
$$\int\limits_{0}^{\infty}\nu(x)\log[\psi_{s}(x\omega)]\ dx=\log[\psi_{s}(\omega)]+O(1).$$
Therefore as \(\omega\rightarrow\infty\)
$$H\bigl{(}\bar{\pi}^{\omega,\nu},{\Pi}\bigr{)}=\log(\omega)+(1+o(1))\log^{*}(\omega).$$
(A.8)
Next, with similar arguments we obtain
$$\sum_{h\in\mathcal{H}}\bar{\pi}^{\omega,\nu}_{h}\biggl{[}H(\bar{\pi}^{\omega,\nu},{\Pi})-\log\frac{1}{{\Pi}_{h}}\biggr{]}_{+}=\int\limits_{0}^{\infty}\nu(x)\Bigl{[}H(\bar{\pi}^{\omega,\nu},{\Pi})-\log(x\omega+1)$$
$${}-(1+o(1))\log[\log(x\omega+1)+1]\Bigr{]}_{+}dx$$
$${}=\int\limits_{0}^{\infty}\nu(x)\Bigl{[}\log(\omega)+(1+o(1))\log^{*}(\omega)-\log(x\omega+1)$$
$${}-(1+o(1))\log[\log(x\omega+1)+1]\Bigr{]}_{+}dx$$
$${}=\int\limits_{0}^{\infty}\nu(x)\bigl{[}-\log(x+1)+o(1)\log^{*}(\omega)\bigr{]}_{+}dx=o(1)\log^{*}(\omega)].$$
(A.9)
In view of (49) and (50) the rest of the proof is similar to the
one of Theorem 3 and therefore omitted.
