Let $X(t)$ be a continuous stochastic process that has increments of fixed length that are strongly stationary, that is $X(t+s)-X(t)$ are identically distributed for any fixed $s>0$. Then I choose $E[X(t+s)-X(t)]=0$ and $v(s)=E[(X(t+s)-X(t))^2]<\infty$ and $\lim_{s\to 0}v(s)=0$. Let the function $f$ be everywhere integrable by Riemann-Stiltjes. My belief is following holds $$ \lim_{h\to 0^+}\int_0^{t}f(X_{s+h})dX_{s+h}=\int_0^{t}f(X_{s})dX_s\tag{1} $$ where the integrals are defined as following sums, with $h$ as step length, $$ \sum_{n=0}^Nf(X_{n+1})(X_{n+2}-X_{n+1})=\sum_{n=0}^Nf(X_n)(X_{n+1}-X_n)\tag{2} $$ which gives $$ f(X_{N+1})(X_{N+1}-X_N)=f(X_0)(X_1-X_0)\tag{3} $$ and is true if $f$ is bounded and we let $h\to 0$.
Intuively I think I could have also used $E[X(t+s)-X(t)]=0$ and that $\lim_{s\to 0}v(s)=0$ which means $\lim_{s\to 0}E[X(t+s)-X(t)]=0$ has no variance. What are additional requirements are needed to make (1) valid?
To make (2) not too trivial I choose the increment $X_{s+h}-X_s$ to be dependent of $X_s$. Is it enough with $f$ being bounded? If yes, then I presume I can just square (3) and use $\lim_{s\to 0}v(s)=0$.
PS: Just for clarification, $X(s)$ is not the Wiener process of function thereof.