When do the Ito integral and the Riemann Stieltjes integral coincide? I have a proof that these two notions coincide on $\mathcal{L}^2_T$, which is clearly not the case (see $\int B_tdB_t$). Can you tell me where I went wrong?
Claim 1. The Ito and the R-S integral coincide on simple processes (i.e. $f(t,\omega)$ is simple if it is bounded and there is a finite partition $\Pi$ of $[0,T]$ such that $f=\sum_{s_i\in\Pi} f(s_{i-1})\mathbb 1_{[s_{i-1},s_i)}$ so all paths of $f$ are simple functions which have a joint partition).
Proof. Let $f$ be a simple process with partition $\Pi$. Then
$$\int f(t) dB_t=\sum_{s_i\in\Pi}f(s_{i-1})(B_{s_{i}}-B_{s_{i-1}})=\omega\mapsto\int f(t,\omega)d B_t(\omega).$$
Here the left side is the Ito integral and the right side is the pointwise R-S integral.
$\square$
Claim 2. The Ito and the R-S integral coincide on $\mathcal{L}^2_T$ (i.e. $L^2$ limits of simple functions).
Proof. Let $f_n\stackrel{L^2}{\to} f$. Then
$$ \int f(t) dB_t =L^2-\lim \int f_n(t) dB_t=L^2-\lim \left[\omega\mapsto\int  f_n(t,\omega) dB_t(\omega)\right].$$
Now $L^2$ convergence implies that there is a subsequence which converges almost surely (also the $L^2$ and the a.s. limit coincide), denote this subsequence again by $f_n$. Therefore
$$\left[\int f(t) dB_t\right](\omega) =\lim \int f_n(t,\omega) dB_t(\omega)\ \ \ \text{almost surely.}$$
Particularly, the a.s. limit of $\int f_n(t) dB_t$ exists (which is not always the case :( ). $\square$ 
 A: I believe the problem here is mainly one of definition. The Ito integral of a process $H\in \mathcal L^2_T$ against $B$ is defined as 
$$\int_0^tH_sdB_s:=\lim_{n\to \infty}\sum_{\substack{i=1 \\s_i\in\pi_n}}^nH_{s_{i-1}}(B_{s_{i}}-B_{s_{i-1}}),$$
where each $\pi_n=\{0\leq s_0^n<\dots<s_n^n\leq t\}$ is a partition of $[0,t]$ such that $\lim_{n\to \infty}mesh(\pi_n)=0$. Furthermore the
limit is taken in probability.
Thus the issue comes in your first claim. Even though $f$ is simple that does not mean that we can take a finite partition and claim that a sum over that finite partition is the integral. Remember Brownian motion has paths of unbounded variation almost surely, so choosing a different partition will result in a completely different process, so your definition in claim 1 is dependent on the partition, and so really cannot be said to be the integral.
Furthermore you cannot fix this by using an infinite sequence of refinements of your chosen partition and taking the pointwise limit, because Brownian motion has paths of unbounded variation almost surely, so this limit won't exist almost surely.
