= Solution
Fix $t>0$, a deterministic <partition of an interval> $\pi$, and write $\eta(s)=t_j$ on $[t_j,t_{j+1})$. Its mesh is $|\pi|=\max_j(t_{j+1}-t_j)$. The <Brownian increment> <variance> gives
$$
\mathbb E\int_0^t|B_s-B_{\eta(s)}|^2ds
=\frac12\sum_j(\Delta t_j)^2\leq\frac{t|\pi|}{2}.
$$
Thus the left-endpoint sums $\sum_jB_{t_j}\Delta B_j$ converge in <L2 space> to the <Itô integral>. Although these step coefficients are unbounded, clipping them at finite levels first makes them bounded <simple predictable processes>; the <Itô isometry> and their finite second <moments> remove the clipping.
The elementary identity $B_{t_{j+1}}^2-B_{t_j}^2=2B_{t_j}\Delta B_j+(\Delta B_j)^2$ gives
$$
2\sum_jB_{t_j}\Delta B_j=B_t^2-\sum_j(\Delta B_j)^2.
$$
The squared <Brownian increments> are <independent>, with <means> $\Delta t_j$ and <variances> $2(\Delta t_j)^2$, since a centered <normal random variable> of <variance> $v$ has fourth <moment> $3v^2$. Therefore
$$
\mathbb E\left|\sum_j(\Delta B_j)^2-t\right|^2
=2\sum_j(\Delta t_j)^2\leq2t|\pi|\longrightarrow0.
$$
Passing to the <convergence in L2> limits yields
$$
\boxed{\int_0^tB_s\,dB_s=\frac12(B_t^2-t).}
$$
This holds for each fixed $t$; continuous <sample paths> of both sides extend equality from rational times to all $t$ on one event of <probability> one.
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