Given a continuous stochastic process $X_{t}$ and a function $f\in C^{1}(\mathbb{R})$ prove that
$\int_{0}^{t}f'(X_{u})dX_{u}=f(X_{t})-f(X_{0})$ Attemp:
I tried using continuity of the $f$ function and the Ito formula but I don't think that I'm right.
We know that $f\in C^{1}(\mathbb{R})$ and that $A_{t}$ is continuous so $f'(A_{t})$ is continuous.That means that $\left|\frac{f(A_{t+h})-f(A_{t})}{A_{t+h}-A_{t}}-f'(A_{t})\right|\longrightarrow0\,\text{since}\,h\rightarrow0\tag{1}$ We know that $\int_{0}^{t}f'(A_{s})dA_{s}=\underset{n\rightarrow\infty}{lim}\mathop{{\displaystyle \sum_{i=0}^{n-1}}}f'(A_{t_{i}})(A_{t_{i+1}}-A_{t_{i}})$
We can rewrite (1):
Assume $\left(t_{i}^{(n)}\right)_{i=0,..,N_{n}}$ a sequence of partitions that
$t_{0}=0<t_{1}<\cdots<t_{n-1}<t_{n}=t$ where $\underset{i\in\mathbb{N}}{sup}|t_{i+1}-t_{i}|\rightarrow0$ since $n\rightarrow\infty$.
then $\forall n\geq n_{0}$ for an $n_{0}\in\mathbb{N} \left|\frac{f(A_{t_{i+1}})-f(A_{t_{i}})}{A_{t_{i+1}}-A_{t_{i}}}-f'(A_{t_{i}})\right|<\varepsilon_{n}$ where $\{\varepsilon_{n}\}_{n\in\mathbb{N}}$ a descending sequence of positive terms with $\varepsilon_{n}<1\,\forall n\in\mathbb{N}$ and $\varepsilon_{n}\rightarrow0$ .
So we have that $\left|f'(A_{t_{i}})\right|\leq\left|f(A_{t_{i+1}})-f(A_{t_{i}})\right|+\varepsilon_{n}\left|A_{t_{i+1}}-A_{t_{i}}\right|\leq\left|f(A_{t_{i+1}})\right|+\left|f(A_{t_{i}})\right|+\varepsilon_{n}\left|A_{t_{i+1}}-A_{t_{i}}\right|$
But f is bounded so $\forall t\in\mathbb{R}$ so $\left|f(A_{t})\right|\leq M$ for an M>0.
$\left|A_{t_{i+1}}-A_{t_{i}}\right|\leq\mathop{{\displaystyle \sum_{i=0}^{n-1}}}\left|A_{t_{i+1}}-A_{t_{i}}\right|$ which is a bounded quantity becasue we have bounded variation.
So $\mathop{{\displaystyle \sum_{i=0}^{n-1}}}\left|A_{t_{i+1}}-A_{t_{i}}\right|\leq N$ for an N>0 so $\left|A_{t_{i+1}}-A_{t_{i}}\right|\leq N$
In addition for the function $f'(A_{t_{i}})(A_{t_{i+1}}-A_{t_{i}}))$ is $\underset{n\rightarrow\infty}{lim}f'(A_{t_{i}})(A_{t_{i+1}}-A_{t_{i}})=\underset{n\rightarrow\infty}{lim}\left(f(A_{t_{i+1}})-f(A_{t_{i}})\right)=f(A_{t_{i+1}})-f(A_{t_{i}})$
and from the theorem of bounded convergence we get that
$\underset{n\rightarrow\infty}{lim}\mathop{{\displaystyle \sum_{i=0}^{n-1}}}f'(A_{t_{i}})(A_{t_{i+1}}-A_{t_{i}})={\displaystyle \sum_{i=0}^{\infty}}\underset{n\rightarrow\infty}{lim}f'(A_{t_{i}})(A_{t_{i+1}}-A_{t_{i}})={\displaystyle \sum_{i=0}^{\infty}}\left(f(A_{t_{i+1}})-f(A_{t_{i}})\right)$
which is a telescoping sum that equals to $f(A_{t_{\infty}})-f(A_{t_{0}})=f(A_{t})-f(A_{0})$ and the conclusion is that
$\int_{0}^{t}f'(A_{s})dA_{s}=f(A_{t})-f(A_{0})$