I am currently working with fractional Brownian motions (fBm) and I want to tho show that $$\lim\limits_{\epsilon\rightarrow 0^+}\int_0^T Y(s)\frac{B(s+\epsilon)-B(s)}{\epsilon}\mathrm{d}s = \lim\limits_{\mu \rightarrow 0}\sum\limits_{j=1}^N Y(t_i)((B(t_{i+1})-B(t_i))$$ Here $Y$ should be a continuous process and $B$ is my fractional brownian motion with Hurts parameter $H>\frac{1}{2}$. Moreover, $\mu$ corresponds to the partition size of the partition $(t_i)$ of $[0,T]$. I initially tried to show convergence in $L^1$ but I failed. My approach was to fix $\epsilon$ and then bound $$\mathbb{E}\left[\left|\int_0^T \frac{1}{\epsilon} Y(s)(B(s+\epsilon)-B(s))\mathrm{d}s - \sum\limits_i Y(t_i)(B(t_{i+1}) - B(t_i)) \right|\right]$$ I tried to squeeze in an additional $Y(t_i)(B(s+\epsilon) - B(s))$ to get
$$ \frac{1}{\epsilon}\mathbb{E} \left[\sum\limits_{i}\int_{t_i}^{t_{i+1}} \left|(Y(s)-Y(t_i) (B(s+\epsilon)-B(s)\right|\mathrm{d}s\right] $$ $$ + \frac{1}{\epsilon}\mathbb{E} \left[ \sum\limits_{i}\int_{t_i}^{t_{i+1}} \left| Y(t_i)(B(s+\epsilon)-B(t_{i+1}) + B(t_i)-B(s)) \right| \mathrm{d}s\right]. $$
But I am unable to bound any of these terms with something of order $\mu$. Especially the $B$ terms in the second summand mess everything up and give me terms that I can neither force to go to zero when $\epsilon$ goes to zero nor $\mu$. Therefore, I tried to approximate $Y$ with a step function $Y^k$ to get something like $$\lim\limits_{\epsilon\rightarrow 0^+}\int_0^T Y^k(s)\frac{B(s+\epsilon)-B(s)}{\epsilon}\mathrm{d}s = \lim\limits_{\epsilon\rightarrow 0^+} \sum\limits_{i} Y^k(s) \int_{t_i}^{t_{i+1}}\frac{B(s+\epsilon)-B(s)}{\epsilon}\mathrm{d}s$$ but I am still unable to show that the second factor equals $B(t_{i+1})-B(t_i)$ as $\epsilon$ goes to zero.
Does anyone have a tip on how to show that?
Best regards any may thanks in advance!