Let $B$ be a Brownian motion. We want to define $$ \int_{0}^{t} B_{s} dB_{s} : = \lim_{n \to \infty } \sum_{k = 1}^{2^{n}t} B_{\frac{k-1}{2^{n}}}[ B_{\frac{k}{2^{n}}} - B_{\frac{k-1}{2^{n}}}]. $$
To do this we notice that $$ B_{t}^{2} =\sum_{k = 1}^{2^{n}t} [ B_{\frac{k}{2^{n}}} - B_{\frac{k-1}{2^{n}}}]^{2} = 2 \sum_{k = 1}^{2^{n}t} B_{\frac{k-1}{2^{n}}}[ B_{\frac{k}{2^{n}}} - B_{\frac{k-1}{2^{n}}}] + \sum_{k=1}^{2^{n}t} [ B_{\frac{k}{2^{n}}} - B_{\frac{k-1}{2^{n}}}]^{2} . $$
Now let $$ S_{n} = \sum_{k=1}^{2^{n}t} [ B_{\frac{k}{2^{n}}} - B_{\frac{k-1}{2^{n}}}]^{2}. $$
From the definition of the Brownian motion, for all $ n $, $ \mathbb{E} [ S_{n} ] = t $ and $ \mathrm{Var}(S_{n}) = 0 $, hence $ S_{n} $ converges in probability to $ t $.
Thus the first limits exists and we get also (unsurprisingly if one knows Ito's formula) that $$ \int_{0}^{t} B_{s} dB_{s} = \frac{B^{2}_{t}}{2} - \frac{t}{2}. $$
Now let $e(t,B_{t'})$ for $t' \leq t$, which we will indicate from now on as $e_{t}$, be a function such that is path wise continuous in $t$ and such that is non-anticipating, i.e., is only depends on Brownian motion up to the time $t$. Notice that this implies that $e_{\frac{k-1}{2^{n}}}$ is independent from $[ B_{\frac{k}{2^{n}}} - B_{\frac{k-1}{2^{n}}}]$.
In the same fashion as before I want to define $$ \int_{0}^{t} e_{s} dB_{s} : = \lim_{n \to \infty } \sum_{k = 1}^{2^{n}t} e_{\frac{k-1}{2^{n}}}[ B_{\frac{k}{2^{n}}} - B_{\frac{k-1}{2^{n}}}]. $$
My question is that I am trying to replicate the proof above in order to prove that the limit exists but everyone of my attempts have failed.
The problem for me is that in this case I do not have the a posteriori knowledge due to Ito's formula that I have used in the previous prove also because I know nothing about the smoothness of $e$.
