Let $B = (B_{t})_{t \geq 0}$ be a 1-dimensional Brownian motion on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ (i.e. that is, $B_t: \Omega \to \mathbb{R}$ for each $t \geq 0$), and let $T > 0$ be a real number. For each $n \in \mathbb{N}$, let
$$0 = t_0^{(n)} < t_1^{(n)} < \cdots < t_{m_n}^{(n)} = T$$
be a partition of $[0,T]$, and suppose that $\max\limits_{0 \leq j \leq m_n-1} |t_{j+1}^{(n)} - t_j^{(n)}| \to 0$ as $n \to \infty$. For each $n \in \mathbb{N}$ define $\phi_n: [0,T] \times \Omega \to \mathbb{R}$ by
$$ \phi_n(t,\omega) := \sum_{j=0}^{m_n-1} B_{t_j^{(n)}}^2 (\omega) \chi_{[t_j^{(n)},t_{j+1}^{(n)})}(t), \quad \forall t \geq 0, \, \omega \in \Omega. $$
I am trying to show the following:
$$ \text{Want To Show:} \qquad \lim_{n \to \infty} \mathbb{E} \left[ \int_{0}^{T} \left(B_t^2(\cdot) - \phi_n(t,\cdot) \right)^2 dt \right] = 0. \hspace{2cm} (1) $$
My attempt. Firstly, we can write $B_t^{2}$ as \begin{align*} B_t^2(\omega) = \sum_{j=0}^{m_n - 1} B_t^2(\omega) \chi_{[t_j^{(n)},t_{j+1}^{(n)})}(t). \end{align*}
Thus, \begin{align*} B_t^2(\omega) - \phi_n(t,\omega) = \sum_{j=0}^{m_n - 1} \left( B_t^2(\omega) - B_{t_j^{(n)}}^2 (\omega) \right) \chi_{[t_j^{(n)},t_{j+1}^{(n)})}(t). \end{align*}
And since the indicator functions have disjoint support, squaring both sides gives
\begin{align*} \left( B_t^2(\omega) - \phi_n(t,\omega) \right)^2 = \sum_{j=0}^{m_n - 1} \left( B_t^2(\omega) - B_{t_j^{(n)}}^2 (\omega) \right)^2 \chi_{[t_j^{(n)},t_{j+1}^{(n)})}(t). \end{align*}
Now integrating both sides from $t = 0$ to $t = T$ and using the linearity of the integral, we obtain \begin{align*} \int_{0}^{T} \left( B_t^2(\omega) - \phi_n(t,\omega) \right)^2 dt = \sum_{j=0}^{m_n - 1} \int_{t_j^{(n)}}^{t_{j+1}^{(n)}} \left( B_t^2(\omega) - B_{t_j^{(n)}}^2 (\omega) \right)^2 dt. \end{align*}
Now taking expectations and applying the linearity of the expectation together with Tonelli's theorem, we obtain \begin{align*} \mathbb{E} \left[ \int_{0}^{T} \left( B_t^2(\cdot) - \phi_n(t,\cdot) \right)^2 dt \right] = \sum_{j=0}^{m_n - 1} \int_{t_j^{(n)}}^{t_{j+1}^{(n)}} \mathbb{E} \left[ \left( B_t^2(\cdot) - B_{t_j^{(n)}}^2 (\cdot) \right)^2 \right] dt. \end{align*}
How should I proceed from here?
I am not sure how to calculate/estimate the last expectation. Since $\{B_t\}_{t \geq 0}$ is a Brownian motion I know that the increments are normally distributed, i.e. $B_t - B_s \sim N(0,t-s)$ for all $0 < s < t$. So I know how to compute the moments of the increments, and I've been trying to express $\left( B_t^2(\omega) - B_{t_j^{(n)}}^2 \right)^2$ in terms of such increments, but so far I have been unsuccessful. Any help would be greatly appreciated.