My question is on the convergence of the Riemann sum, when the value spaces are square-integrable random variables. The convergence does depend on the evaluation point we choose, why is the case. Here is some background to make this clearer.
Suppose $f\colon \Re \mapsto \Re $ is some continuous function on $[a,b]$, the Stieltjes integral of $f$ with respect to itself $f$ is $\int^{b}_{a} f(t)df(t)$
if we take a partition $ \Delta_n = \{t_0, t_1, \cdots, t_n \}$ of $[a,b]$ the Riemmans sums is
$$ L_{n} = \sum^{n}_{i=1} f(t_{i-1})(f(t_{i})-f(t_{i-1})) $$
Now if the limit exists say $\lim \limits_{n\to\infty} L_{n}= A$, then if we choose the evaluation point $t_{i}$ then the sum
$$ R_{n} = \sum^{n}_{i=1} f(t_{i})(f(t_{i})-f(t_{i-1})) $$
will also converge to $A$ so
$$\lim_{n\to\infty}L_{n} = \lim_{n\to\infty}R_{n} .$$
Now we apply same idea for a stochastic integral. Here $W(t)$ is a wiener process and we wish to find
$$\int^{b}_{a}W(t)dW(t) $$
$$ L_{n} = \sum^{n}_{i=1} W(t_{i-1})(W(t_{i})-W(t_{i-1})) $$
$$ R_{n} = \sum^{n}_{i=1} W(t_{i})(W(t_{i})-W(t_{i-1})) $$
in $L^2$ norm the limits of $L_{n}$ and $R_{n}$ exist but are different
$$\lim_{n\to\infty} \Vert R_{n}-L_{n}\Vert = b-a $$
can someone explain why the limits are different ? If the limit exists which in this case it does. I would have expected $\lim_{n\to\infty} \Vert R_{n}-L_{n}\Vert = 0 $ in $L^2$ norm.
