Given a partition of the interval $［0,T］$ by setting $t_k=k\Delta t$, $k=0,...,K-1$ and $K\Delta t=T$ and a parameter $\lambda\in[0,1]$ and set$$\tau_k=(1-\lambda)t_k+\lambda t_{k+1}.$$ I am looking for proof for following proposition:
Assume that there exist $C$, $\delta>0$ such that $\mathbb E[(f(t)-f(s))^2]\leq C|t-s|^{1+\delta}, 0\leq s,t\leq T$. Then the Riemann-sum approximation$$I(t):=\lim_{K\rightarrow\infty}\sum_{k=0}^{K-1}f(\tau_k)\left(W_{t_{k+1}} -W_{t_k})\right)$$ converges in $L^1(\Omega)$ to the same value for all $\lambda\in[0,1]$
How can I proof this? I thought the difference between different point $\tau_k$ will give different value almost surely as the variation is not finite.
Write
$$\sum_{k=0}^{K-1} f(\tau_k) (W_{t_{k+1}}-W_{t_k}) = I_1 + I_2$$
where
$$\begin{align*} I_1 &:= \sum_{k=0}^{K-1} f(t_k) (W_{t_{k+1}}-W_{t_k}) \\ I_2 &:= \sum_{k=0}^{K-1} (f(\tau_k)-f(t_k)) (W_{t_{k+1}}-W_{t_k}). \end{align*}$$
It is well-known that $I_1 \xrightarrow[]{K \to \infty} \int_0^T f(s) \, dB_s$ in $L^2$ (hence in $L^1$). We are done if we can show that $I_2 \to 0$ in $L^1$. To this end, we note that
$$\mathbb{E}(|I_2|) \leq \sum_{k=0}^{K-1} \mathbb{E}\bigg[ |(f(\tau_k)-f(t_k)) (W_{t_{k+1}}-W_{t_k})| \bigg].$$
Applying Cauchy's inequality, we find
$$\mathbb{E}(|I_2|) \leq \sum_{k=0}^{K-1} \sqrt{\mathbb{E}((f(\tau_k)-f(t_k))^2)} \sqrt{\mathbb{E}((W_{t_{k+1}}-W_{t_k})^2)}$$
and therefore
$$\mathbb{E}(|I_2|) \leq C \sum_{k=0}^{K-1} (t_{k+1}-t_k)^{1+\delta/2} \leq T \max_k |t_{k+1}-t_k|^{\delta/2} \xrightarrow[]{K \to \infty} 0.$$