\int^{\cdot}_0 h^k (s) dW^k_s)(t)$$ holds for all $t\leq T$ (a.s.) and also in $L_2(\Omega\times [0,T])$.

[(ii)]{} Suppose $
 h_n
 =(h^1_n,h^2_n,\cdots)$ converges to $h=(h^1,h^2,\cdots)$ in $L_2(\Omega\times [0,T],{\mathcal{P}}, \ell_2)$ as $n\to \infty$. Then, as $n\to \infty$, $$\label{eqn 8.30.2}
 \sum_{k=1}^\infty I^{a}\int^{\cdot}_0 h^k_n dW^k_s  \quad
  \hbox{converges to} \quad  \sum_{k=1}^\infty I^{a}\int^{\cdot}_0 h^k (s) dW^k_s$$ in probability uniformly on $[0,T]$.

Since the series $\sum_{k=1}^{\infty} \int^t_0 h^k (s) dW^k_t$ converges in $L_2(\Omega \times [0,T])$, by (\[eqn 7.03.1\]) we have $$\begin{aligned}
I^{a}\Big( \sum_{k=1}^{\infty} \int^{\cdot}_0 h^k (s) dW^k_s \Big)(t)&=&\lim_{n\to \infty} I^{a}\Big( \sum_{k=1}^{n} \int^{\cdot}_0 h^k (s) dW^k_s \Big)(t) \\
&=&\lim_{n\to \infty} \sum_{k=1}^{n} I^{a} \Big( \int^{\cdot}_0 h^k (s) dW^k_s \Big)(t) \quad \quad \text{in}\,\,\,L_2(\Omega \times [0,T]).\end{aligned}$$ Thus, the series $\sum_{k=1}^{\infty} I^{a} \int^t_0 h^k (s) dW^k_s$ converges in $L_2(\Omega\times [0,T])$ and $$\label{eqn 7.04.1}
I^{a}\sum_{k=1}^{\infty} \int^t_0 h^k (s) dW^k_s=\sum_{k=1}^{\infty} I^{a} \int^t_0 h^k (s) dW^k_s$$ in $L_2(\Omega\times [0,T])$, and thus the equality holds (a.e.). Also by Burkholder-Davis-Gundy inequality $$\begin{aligned}
&&{\mathbb{E}}\Big[ \sup_{t\leq T} \Big|\sum_{k=n}^{m} I^{a} \int^t_0 h^k (s) dW^k_s
\Big|^2 \Big] \nonumber\\
&\leq& N{\mathbb{E}}\Big[ \sup_{t\leq T} \Big|\sum_{k=n}^{m}  \int^t_0 h^k (s) dW^k_s
\Big|^2 \Big] \leq N \sum_{k=n}^{m} {\mathbb{E}}\Big[ \int^T_0    |h^k(s)|^2 ds \Big]
\to 0 \label{eqn 8.29.1}\end{aligned}$$ as $n,m\to \infty$. Therefore the series $\sum_{k=1}^{\infty} I^{a} \int^t_0 h^k (s) dW^k_s$ converges in probability uniformly on $[0,T]$. It follows that 