Next, for $\mathcal{N}_{35},$ we have $$\begin{aligned}
    \mathcal{N}_{35}
    &\leq&
    C\mathbb{E} \Big[\sup _{t \in[0, T]} \Big(\Lambda^{0,T}_{\alpha,B^{H,N,n,m}}\int_{0}^t(r^{-\alpha}+(t-r)^{-\alpha})(1+|u^{N,n}_r|)\mathrm{d}r\mathbf{1}_{D_{t}^{N,R}}
    \Big)^2\Big]\cr
    &&+
    C\mathbb{E}\Big[\sup _{t \in[0, T]} \Big(\Lambda^{0,T}_{\alpha,B^{H,N,n,m}}\int_{0}^{t}\Big(\int_{0}^{r}\frac{|u^{N,n}_r-u^{N,n}_q|}{(r-q)^{1+\alpha}}\mathrm{d}q\Big)\mathrm{d}r\mathbf{1}_{D_{t}^{N,R}}
    \Big)^2\Big]\cr
    &\leq & C\mathbb{E} \Big[ \int_{0}^{T}\big(\Lambda^{0,T}_{\alpha,B^{H,N,n,m}}\big)^2(1+\|u^{N,n}\|^2_{\alpha, t})\mathbf{1}_{D_{t}^{N,R}}\mathrm{d}t\Big]\cr
    &&+ C \mathbb{E} \Big[\big(\Lambda^{0,T}_{\alpha,B^{H,N,n,m}}\big)^2 \int_{0}^{T}\Big(\int_{0}^{t}\frac{|u^{N,n}_t-u^{N,n}_s|}{(t-s)^{1+\alpha}}\mathrm{d}s\Big)^2\mathbf{1}_{D_{t}^{N,R}}\mathrm{d}t\Big]\cr
    &\leq & C_{R}\mathbb{E} \big[\big(\Lambda^{0,T}_{\alpha,B^{H,N,n,m}}\big)^2\big].\end{aligned}$$

In the same way as for the term $\mathcal{M}_{42}$ in the proof of Lemma \[lemboundN\], we have $$\begin{aligned}
    \mathcal{N}_{36}&\leq& 
    C_{N} \mathbb{E}\Big[\sup _{t \in[0, T]} \Big(\int_{0}^t(r^{-\alpha}+(t-r)^{-\alpha})|u^{N,n}_r-u^{N,m}_r|\mathrm{d}r
    \Big)^2\Big]\cr
    &&+C_{N}  \mathbb{E}\Big[\sup_{t\in[0,T]} \Big(\int_{0}^t\Big(\int_{0}^{r}\frac{|u^{N,n}_r-u^{N,m}_r-u^{N,n}_q+u^{N,m}_q|}{(r-q)^{1+\alpha}}\mathrm{d}q\Big)\mathrm{d}r\mathbf{1}_{D_{t}^{N,R}}\Big)^2\Big]\cr
    &&+C_{N}  \mathbb{E}\Big[\sup_{t\in[0,T]}\Big(\int_{0}^t |u^{N,n}_r-u^{N,m}_r|\Big(\int_{0}^{r}\frac{|u^{N,n}_r-u^{N,n}_q|}{(r-q)^{1+\alpha}}\mathrm{d}q\Big)\mathrm{d}r\mathbf{1}_{D_{t}^{N,R}}\Big)^2\Big]\cr
    &&+C_{N}  \mathbb{E}\Big[\sup_{t\in[0,T]}\Big(\int_{0}^t  |u^{N,n}_r-u^{N,m}_r|\Big(\int_{0}^{r}\frac{|u^{N,m}_r-u^{N,m}_q|}{(r-q)^{1+\alpha}}\mathrm{d}q\Big)\mathrm{d}r\mathbf{1}_{D_{t}^{N,R}}\Big)^2\Big]\cr
    &\leq& C_{N,R}  \int_{0}^{T} \mathbb{E}\big[\|u^{N,n}-u^{N,m}\|^2_{\alpha,t}\mathbf{1}_{D_{t}^{N,R}}\big]\mathrm{d}t\cr
    &&+C_{N,R} \mathbb{E}\Big[\int_{0}^T \Big(\int_{0}^{t}\frac{|u^{N,n}_t-u^{N,m}_t-u^{N,n}_s+u^{N,m}_s|}{(t-s)^{1+\alpha}}\mathrm{d}s\mathbf{1}_{D_{t}^{N,R}}\Big)^2\mathrm{d}t\Big]\cr
    &\leq& C_{N,R}  \int_{0}^{T} \mathbb{E}\big[\|u^{N,n}-u^{N,m}\|^2_{\alpha,t}\mathbf{1}_{D_{t}^{N,R}}\big]\mathrm{d}t.\end{aligned}$$

Finally, by Gronwall’s lemma, we have $$\begin{aligned}
    \mathbb{E}\Big[\|u^{N,n}-u^{N,m}\|^2_{\alpha,T}\mathbf{1}_{D_{T}^{N,R}}\Big] \leq C_{N,R} \mathbb{E}\Big[\Big(\sum_{i=1}^{\infty}\sqrt{\lambda_{i}}\|\beta_{i}^{H,N,n}-\beta_{i}^{H,N,m }\|_{\alpha,0,T}\Big)^2\Big].\end{aligned}$$ This completes the proof of Lemma \[lemboundNm\].

For fixed $N\geq 1$, we will show that the sequence $\{u^{N,n}, n\geq 1\}$ is fundamental in probability in the norm $\|\cdot\|_{\alpha,T}$. For all $\varepsilon>0$, $R\geq 1$, we have $$\begin{aligned}
\mathbb{P}\big(\|u^{N,n}-u^{N,m}\|_{\alpha,T}>\varepsilon\big) &\leq&
\mathbb{P}\big( \|u^{N,n}\|_{\alpha,T} > R~{\rm or}~ \|u^{N,m}\|_{\alpha,T} > R \big) \cr
&&+\mathbb{P}\big(\|u^{N,n}-u^{N,m}\|_{\alpha,T}>\varepsilon, \|u^{N,n}\|_{\alpha,T} \leq R, \|u^{N,m}\|_{\alpha,T} \leq R \big).\end{aligned}$$

Since $\varpi \in(1-\alpha, H)$, by (\[betaNn-N\]), we have $$\begin{aligned}
\label{boundbetan}
\sup_{i\in \mathbb{N}}\mathbb{E}\big[\|\beta_{i}^{H,N,n}-\beta_{i}^{H,N}\|^2_{\alpha,0,T}\big] &\leq& C(\frac{1}{n})^{2(\varpi+\alpha-1)} \sup_{i\in \mathbb{N}} \mathbb{E}\big[\|\beta_{i}^{H}\|^2_{1-\varpi,0,T}\big]\cr
&\leq& C(\frac{1}{n})^{2(\varpi+\alpha-1)}.\end{aligned}$$ where $C$ is a constant which is independent of $i$ and $N$.

Then, by Cauchy-Schwarz’s inequality, we have $$\begin{aligned}
\mathbb{E}\Big[\Big(\sum_{i=1}^{\infty}\sqrt{\lambda_{i}}\|\beta_{i}^{H,N,n}-\beta_{i}^{H,N}\|_{\alpha,0,T}\Big)^2\Big]
&=&\mathbb{E}\Big[\Big(\sum_{i=1}^{\infty}\sqrt[4]{\lambda_{i}}\sqrt[4]{\lambda_{i}}\|\beta_{i}^{H,N,n}-\beta_{i}^{H,N}\|_{\alpha,0,T}\Big)^2\Big]\cr
&\leq& \sum_{i=1}^{\infty}\sqrt{\lambda_{i}}\Big(\sum_{i=1}^{\infty}\sqrt{\lambda_{i}}\mathbb{E}\big[\|\beta_{i}^{H,N,n}-\beta_{i}^{H,N}\|^2_{\alpha,0,T}\big]\Big).\end{aligned}$$

Thus, the condition $\sum_{i=1}^{n}\sqrt{\lambda_{i}}<\infty$ and (\[boundbetan\]) yield 