Let $dX_t=\mu(t,\mu)dt+\sigma (t,\mu)dB_t$. If $f\in C^2(\mathbb R)$, then $$df(X_t)=f'(X_t)dX_t+\frac{1}{2}f''(X_t)dt.$$
The proof goes as follow. Let $0=t_0<t_1<...<t_n=t$ a partition of $[0,t]$ s.t. $\pi_n=\max_{i=0,...,n-1}|\Delta t_i|\to 0$ when $n\to \infty $. I denote $$\Delta X_i=X_{t_{i+1}}-X_{t_i},\ \ \Delta t_i=t_{i+1}-t_i.$$ Since $f\in C^2(\mathbb R)$, $$f(X_t)=\sum_{i=0}^{n-1}f(X_{i+1})-f(X_i)=\sum_{i=0}^{n-1}f'(X_i)\Delta X_i+\sum_{i=0}^{n-1}R_i(\Delta X_i),$$ where $R_i(x)=o(x^2)$. Then, in the book of Oksendal (SDE edition 6) they say that $\sum_{i=0}^{n-1}R_i\to 0$ when $n\to \infty $.
Q1) This is a mistake and should be $\sum_{i=0}^{n-1}R_i\to 0$ in $L^2$ when $n\to \infty $, no ? Because this is not true pointwise, no ?
So, I tried to prove $$\sum_{i=0}^{n-1}R_j(\Delta X_i)=\sum_{i=0}^{n-1}\varepsilon _i(\Delta X_i)(\Delta X_i)^2,$$
Since $\varepsilon _i(x)\to 0$ when $x\to 0$, we know that there exist $\delta $ s.t. $|x|\leq \delta $ implies $|\varepsilon _i(x)|<1$ for all $i=0,...,n$.
Now, since $|\Delta X_i|<\delta $ for all $i$ is not true, I'm not so sure why $\varepsilon _i(\Delta X_i)<1$ for all $i$ and thus why $\varepsilon _i(\Delta X_i)\in L^1$.
Q2) Any idea to conclude ?