A martingale $\{u_n\}$ is $\mathcal{L}^2-bounded$. Show that: $$ \lim \int (u-u_n)^2 d\mu = 0 $$
$$ \int (u_{j+k} -u_j)^2 d\mu \! \begin{aligned}[t] & = \sum_{l=j+1}^{j+k} \int (u_l-u_{l-1})^2 d\mu \end{aligned} $$ Now, since it's positive and measurable: $$ \int \liminf(u-u_n)^2 d\mu \! \begin{aligned}[t] & \leq \liminf \int (u-u_n)^2 d\mu \\ & \leq \limsup \int (u-u_n)^2 d\mu \\ & \leq \limsup\sum_{l= n+1}^{\infty} \int (u_l-u_{l-1})^2 d\mu \\ & = 0 \end{aligned} $$
The book reasoned that zero due to the $\mathcal{L}^2$-boundedness of $\sum_{k=1}^{\infty} \int (u_k-u_{k-1})^2 d\mu <\infty$.
Could someone explain me why the last result is zero?
It should read $$ \int \liminf(u-u_n)^2 d\mu \! \begin{aligned}[t] & \leq \liminf \int (u-u_n)^2 d\mu \\ & \leq \limsup \int (u-u_n)^2 d\mu \\ & \leq \limsup\sum_{l=n+1}^{\infty} \int (u_l-u_{l-1})^2 d\mu \\ & = 0 \end{aligned} $$ It is an elementary fact that if $a_n \geq 0$ and $\sum a_n <\infty$ then $\sum_{l=n+1}^{\infty} a_l \to 0$.