in noncommutative $L_p(M,\Phi)$ spaces.

Let $\mathbf{x}=\{x_n\}$ be an $L_p(M,\Phi)$-martingale and $\{w_n\}_{n\geq1}$ be a sequence of positive numbers such that $W_n=\sum\limits_{k=1}^nw_k\rightarrow\infty$ as $n\rightarrow\infty$. *The weighted average* (see [@ki1],[@ZH]) $\sigma_n(\mathbf{x})$ of the martingale $x=\{x_n\}$ is given by $$\sigma_n(\mathbf{x})=\frac{1}{W_n}\sum\limits_{k=1}^nw_k x_k.$$

\[3.8\] Let $\mathbf{x}=\{x_n\}$ be an $L_p(M,\Phi)$-martingale and $\sigma_n(\mathbf{x})$ be a weighted average of $\mathbf{x}$. Then one has $$\sup\limits_{n\geq 1}\|x_n\|_p=\sup\limits_{n\geq 1}\|\sigma_n(\mathbf{x})\|_p.$$

Due to $$(\sup\limits_{n\geq 1}\|x_n\|_p)(\omega)=\sup\limits_{n\geq
1}\|x_n(\omega)\|_{L_p(M(\omega),\tau_\omega)}$$ for almost all $\omega\in \Omega$, from Theorem 3.2 [@ZH] we obtain $$\begin{aligned}
(\sup\limits_{n\geq 1}\|x_n\|_p)(\omega)&=&\sup\limits_{n\geq
1}\|x_n(\omega)\|_{L_p(M(\omega),\tau_\omega)}\\[2mm]
&=&\sup\limits_{n\geq
1}\|\sigma_n(\mathbf{x}(\omega))\|_{L_p(M(\omega),\tau_\omega)}\\[2mm]
&=&\sup\limits_{n\geq 1}\|\sigma_n(\mathbf{x})\|_p(\omega)\end{aligned}$$ for almost all $\omega\in \Omega$, here we have used the notation $\mathbf{x}(\omega))=\{x_n(\w)\}$. Hence, we obtain the required equality.

\[3.9\] Let $\mathbf{x}=\{x_n\}$ be an $L_p(M,\Phi)$-martingale and $\sigma_n(x)$ be a weighted average of $\mathbf{x}$. Then one has $\{x_n\}_{n\geq1}$ converges in $L_p(M,\Phi)$ if and only if $\sigma_n(\mathbf{x})$ converges in $L_p(M,\Phi)$.

Let $x_n\rightarrow y$ in $L_p(M,\Phi)$. Then $x_n(\omega)\rightarrow y(\omega)$ in $L_p(M(\omega),\tau_\omega)$ for almost all $\omega\in
\Omega$. According to Theorem 3.7 [@ZH] $\sigma_n(\mathbf{x}(\omega))\rightarrow y(\omega)$ for almost all $\omega\in \Omega$. Then $\sigma_n(\mathbf{x})(\omega)=\sigma_n(\mathbf{x}(\omega))\rightarrow
y(\omega)$ in $L_p(M(\omega),\tau_\omega)$ for almost all $\omega\in
\Omega$. Hence $$\|\sigma_n(\mathbf{x})-y\|_p(\omega)=\|\sigma_n(\mathbf{x}(\omega))-y(\omega)\|_{L_p(M(\omega),\tau_\omega)}\rightarrow
0$$ for almost all $\omega\in \Omega$, which means $\sigma_n(\mathbf{x})$ converges in $L_p(M,\Phi)$.

Let $\sigma_n(\mathbf{x})\rightarrow y$ in $L_p(M,\Phi)$. Then $\sigma_n(\mathbf{x}(\omega))\rightarrow y(\omega)$ in $L_p(M(\omega),\tau_\omega)$ for almost all $\omega\in \Omega$. Again by Theorem 3.7 [@ZH] $x_n(\omega)\rightarrow 