&   & \beta(\Delta(a,\pi_{-(t-s)}(m)),N_{i}(s-a,\Delta(a,\pi_{-(t-s)}(m))))da \Big) K(t-s,m) ds,  \\
\end{array}$$ for $t \geq t_{n}+\overline{\tau}$. Since $$t-t_{n}-\overline{\tau} \geq \ln\bigg(\frac{h(b_{n+1})}{h(b_{n})}\bigg),$$ we have $$e^{-(t-t_{n}-\overline{\tau})} \leq \frac{h(b_{n})}{h(b_{n+1})}.$$ Hence $$h(m)e^{-(t-t_{n}-\overline{\tau})} \leq h(m)\frac{h(b_{n})}{h(b_{n+1})}\leq h(b_{n+1})\frac{h(b_{n})}{h(b_{n+1})}=h(b_{n}),$$ because $h$ is an increasing function. Consequently, $$\pi_{-(t-t_{n}-\overline{\tau})}(m)=h^{-1}(h(m)e^{-(t-t_{n}-\overline{\tau})}) \leq b_{n}.$$ Using the induction hypothesis, we obtain that $$N_{1}(t_{n}+\overline{\tau},\pi_{-(t-t_{n}-\overline{\tau})}(m))=N_{2}(t_{n}+\overline{\tau},\pi_{-(t-t_{n}-\overline{\tau})}(m)).$$ Finally, since $s-a \geq t_{n}$ and $$\Delta(a,\pi_{-(t-s)}(m)) \leq \Delta(a,m) \leq \Delta(\underline{\tau},m) \leq \Delta(\underline{\tau},b_{n+1})= b_{n},$$ we obtain that $$|N_{1}(t,m)-N_{2}(t,m)| = |J(N_{1})(t,m)-J(N_{2})(t,m)|.$$ Using the Gronwall’s lemma, we obtain the equality for $n+1$.

Noting that there exists $M \in \mathbb{N}^{*}$ such that $b_{M} < g(1)=b_{M+1}$, we obtain, for all $m \in [0,b_{M+1}]=[0,1]$ and $t\geq t_{M+1}$, that $N_{1}(t,m)=N_{2}(t,m)$.

We now suppose that $b\in(h^{-1}(e^{-\underline{\tau}}),g(1)]$. Then the above result is still valid with $M=0$. Thus, $$\label{property2}
N_1(t,m)=N_2(t,m) \qquad \textrm{ for } m\in[0,g(1)] \textrm{ and } t\geq t_{M+1}, \
M\in\mathbb{N}.$$

We finally suppose that $b\in(g(1),1]$.

Let $m\in[g(1),1]$ and $t\geq t_{M+1}+\\overline{\tau}$ be given. For $s\in[t_{M+1}+\overline{\tau},t]$ and $a\in[\underline{\tau},\overline{\tau}]$, we have $s-a\geq
t_{M+1}$. Thus, if $\chi(-(t-s),m)\leq g(1)$, $$\Delta(a,\chi(-(t-s),m))\leq \Delta(a,g(1))\leq h^{-1}(e^{-\underline{\tau}})<g(1)$$ and, from (\[property2\]), $$N_{1}(s-a,\Delta(a,\chi(-(t-s),m)))=N_{2}(s-a,\Delta(a,\chi(-(t-s),m))),$$ and if $\chi(-(t-s),m)>g(1)$ then, from the definition of $\zeta$, $$\zeta(\chi(-(t-s),m),a)=0.$$ Moreover, since $h(m)\leq 1$ for $m\in[0,1]$, we obtain, for $m\in[g(1),1]$ and $t\geq
t_{M+1}+\overline{\tau}-\ln(h(g(1)))$, $$\chi(-(t-t_{M+1}-\overline{\tau}),m)\leq h^{-1}(h(m)h(g(1))) \leq g(1).$$ We deduce that $$N_{1}\big(t_{M+1}+\overline{\tau},\chi(-(t-t_{M+1}-\overline{\tau}),m)\big)=N_{2}\big(t_{M+1}+\overline{\tau},\chi(-(t-t_{M+1}-\overline{\tau}),m)\big).$$ Using once again the Gronwall’s Lemma, we get $$N_{1}(t,m)=N_2(t,m) \qquad \textrm{ 