Approximation of stochastic integral Let $f \in C^2_C(\mathbb{R})$ and $$X_t = X_0 + \int_0^t \sigma(s) \, dB_s + \int_0^t b(s) \, ds$$ be an (one-dimensional) Itô process where $\sigma,b: [0,\infty) \times \Omega \to \mathbb{R}$ progressively measurable and pathwise bounded. Let $(\tau_n)_n$ be a common localization sequence of $b$, $\sigma$. Then there exist sequences of simple processes $(\sigma^{\Pi})_{\Pi}$, $(b^{\Pi})_{\Pi}$ such that $$\sigma^{\Pi} \cdot 1_{[0,\tau_n)} \stackrel{L^2(\lambda_T \times \mathbb{P})}{\to} \sigma \cdot 1_{[0,\tau_n)} \qquad (|\Pi| \to 0) \\ b^{\Pi} \cdot 1_{[0,\tau_n)} \stackrel{L^2(\lambda_T \times \mathbb{P})}{\to}b \cdot 1_{[0,\tau_n)} \qquad (|\Pi| \to 0)$$
for all $n \in \mathbb{N}$. Denote by $X^{\Pi}$ the corresponding Itô process. Then one can actually show that $$\int_0^{t} \sigma^{\Pi}(s) \, dB_s \to \int_0^t \sigma(s) \, dB_s \qquad \quad \int_0^t b^{\Pi}(s) \, ds \to \int_0^t b(s) \, ds \tag{1}$$ as $|\Pi| \to 0$ where the limits are uniform in probability. Thus in particular, $X^{\Pi} \to X$ uniform in probability as $|\Pi| \to 0$
In a proof of Itô's formula for Itô processes,
$$f(X_t)-f(X_0)= \int_0^t f'(X_s) \sigma(s) \, dB_s + \int_0^t f'(X_s)  b(s) \, ds + \frac{1}{2} \int_0^t f''(X_s) \sigma^2(s) \, ds \tag{2}$$
the author claims that it suffices to prove the formula for Itô processes where $b$, $\sigma$ are simple processes, because one can approximate (uniform in probability) an arbitrary Itô process by such processes, as already mentioned.
I don't see why the right-hand side in $(2)$ converges also uniform in probability, i.e. why $$\mathbb{P} \left( \sup_{0 \leq t \leq T} \left| \int_0^t f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) \, dB_s - \int_0^t f'(X_s) \cdot \sigma(s) \, dB_s \right| > \varepsilon \right) \to 0 \tag{3}$$ 
as $|\Pi| \to 0$. It looks similar to the statement in $(1)$, but there I can apply the maximum inequality and Itô isometry since I know that $\sigma^{\Pi} \cdot 1_{[0,\tau_n)} \to \sigma \cdot 1_{[0,\tau_n)}$ in $L^2(\lambda_T \times \mathbb{P})$. Whereas in this case, I have $f'(X_s^{\Pi}) \to f'(X_s)$ uniform in probability, but as far as I can see no convergence in $L^2(\lambda_T \times \mathbb{P})$. How to conclude $(3)$...?
Any hints would be appreciated.
 A: Actually, the proof is indeed similar to the proof of $(1)$. It's based on the fact that convergence in probability implies almost sure convergence of a subsequence. Set 
$$\tau_n := \inf\left\{t \geq 0; \int_0^t |\sigma(s)|^2 \, ds + \int_0^t |b(s)|^2 \, ds \geq n \right\}.\tag{4}$$
By Doob's inequality, Itô's isometry and Tschbysheff inequality we have
$$\begin{align} & \quad \mathbb{P} \left( \sup_{t \leq T} \left| \int_0^t (f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s)) \, dB_s \right| > \varepsilon \right) \\ &\leq \mathbb{P} \left( \sup_{t \leq T} \left| \int_0^{t \wedge \tau_n} (f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s)) \, dB_s \right| > \varepsilon, \tau_n > T \right) + \mathbb{P}(\tau_n \leq T) \\ &\leq \frac{4}{\varepsilon^2} \cdot \underbrace{\mathbb{E} \left( \int_0^{T \wedge \tau_n} |f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s)|^2 \, ds \right)}_{=:I} + \mathbb{P}(\tau_n \leq T) \end{align} $$
(Note that the boundedness of $f'$ implies the existence of the integrals.) Since \begin{align*} f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s) &= f'(X_s^{\Pi}) \cdot \big(\sigma^{\Pi}(s)-\sigma(s)\big) \\ &\quad + \sigma(s) \cdot \big(f'(X_s^{\Pi})-f'(X_s) \big)\end{align*} and $(a+b)^2 \leq 2a^2+2b^2$ we obtain $$\begin{align*} I &\leq 2 \mathbb{E} \bigg( \int_0^{T \wedge \tau_n} \underbrace{|f'(X_s^{\Pi})|^2}_{\leq \|f'\|^2_{\infty}} \cdot |\sigma^{\Pi}(s)-\sigma(s)|^2 \, ds \bigg)\\ &\quad + 2 \mathbb{E} \left( \int_0^{T \wedge \tau_n} \sigma^2(s) \cdot |f'(X_s^{\Pi})-f'(X_s)|^2 \, ds \right) \end{align*}$$
The first addend converges to $0$ as $|\Pi| \downarrow 0$ since $\sigma^{\Pi} \cdot 1_{[0,\tau_n)} \to \sigma \cdot 1_{[0,\tau_n)}$ in $L^2(\lambda_T \times \mathbb{P})$ by assumption. For the second one, we note that 
\begin{align*} \int_0^{T \wedge \tau_n} \sigma^2(s) |f'(X_s^{\Pi})-f'(X_s))|^2 \, ds &\leq \sup_{s \leq T}|f'(X_s^{\Pi})-f'(X_s))|^2  \int_0^{\tau_n} |\sigma(s)|^2 \, ds \\ &\leq n \sup_{s \leq T}|f'(X_s^{\Pi})-f'(X_s))|^2 \end{align*}
Since $X^{\Pi} \to X$ uniformly in probability, it follows from the continuity of $f'$ that the right-hand side converges to $0$ in probability as $|\Pi| \to 0$. Moreover,  by the above estimate,
$$\int_0^{T \wedge \tau_n} \sigma^2(s) |f'(X_s^{\Pi})-f'(X_s))|^2 \,d s \leq 2n \|f'\|_{\infty} \in L^1(\mathbb{P})$$
and so Vitali's convergence theorem gives
$$ \mathbb{E} \left( \int_0^{T \wedge \tau_n} \sigma^2(s) \cdot |f'(X_s^{\Pi})-f'(X_s)|^2 \, ds \right)  \to 0$$
Similarily, one can prove the convergence of the other addends in the right-hand side of $(2)$.
A: So here we assume that $f$ is compactly supported and of $C^2$.
I will prove $$P(\sup_{0\le t \le T} | \int_0^t f''(X_s^\Pi) \sigma_\Pi^2(s)ds - \int_0^t f''(X_s)\sigma^2(s)ds|>\epsilon) \to 0.$$
The initial step is identical as in the first case by decomposing into sets of $\tau_n>T$ and $\tau_n\le T$. So consider the bound $P(\sup_{0\le t \le T} | \int_0^{t \wedge \tau_n} f''(X_s^\Pi) \sigma_\Pi^2(s) - f''(X_s)\sigma^2(s)ds|>\epsilon)\le P(\int_0^{T \wedge \tau_n}|f''(X_s^\Pi)\sigma_\Pi^2(s)-f''(X_s)\sigma^2(s)|ds>\epsilon).$
We take the localizing sequence in as the above answer, i.e.
$\tau_n=\inf \{t\ge 0: \int_0^t \sigma^2(s)ds + \int_0^t b^2(s) ds \ge n\}$. Then since we have by assumption, $\sigma_\Pi 1_{[0,\tau_n)} \to \sigma1_{[0,\tau_n)}$ in $L^2(\lambda_T \times P)$, we have $\sigma_\Pi^21_{[0,\tau_n)} \to \sigma^21_{[0,\tau_n)}$ in $L^1(\lambda_T \times P)$. Using Markov's inequality, we can bound the above by $\frac{1}{\epsilon}E\int_0^{T \wedge \tau_n}|f''(X_s^\Pi)\sigma_\Pi^2(s)-f''(X_s)\sigma^2(s)|ds$.
Now decompose the above to $\frac{1}{\epsilon}E\int_0^{T \wedge \tau_n}|f''(X_s^\Pi)(\sigma_\Pi^2(s)-\sigma^2(s))|ds+\frac{1}{\epsilon}E\int_0^{T \wedge \tau_n}|f''(X_s^\Pi)-f''(X_s)|\sigma^2(s)ds$.
The first integral is bounded by $||f''||_\infty \frac{1}{\epsilon}E\int_0^T 1_{[0,\tau_n)}|\sigma_\Pi^2(s)-\sigma^2(s)|ds$ and this $\to0$ as $|\Pi|\to 0$, since we have $\sigma_\Pi^21_{[0,\tau_n)} \to \sigma^21_{[0,\tau_n)}$ in $L^1(\lambda_T \times P)$.
For the next integral, we use  $\int_0^t \sigma^2(s)1_{[0,\tau_n)} \le n^2$ for all $s\le t$, and the uniform convergence of $X_s^\Pi$ to $X_s$ in probability and uniform continuity of $f''$ to conclude that
$f''(X_s^\Pi)$ converges to $f''(X_s)$ uniformly in probability as well. Also, we have $\int_0^{T \wedge \tau_n} |f''(X_s^\Pi) - f''(X_s)| \sigma^2(s)ds \le \sup_{s\le T} |f''(X_s^\Pi) -f''(X_s)| n$ and $\int_0^T|\sigma^2(s)(f''(X_s^\Pi)-f''(X_s))|1_{[0,\tau_n)} \le Tn4||f''||_\infty^2$ for almost all $\omega$ and $s\le T$, so by Vitali's convergence theorem which generalizes dominated convergence theorem to the case of convergence in probability, we get the second integral $\to 0$ as $|\Pi| \to 0$ as well.
The proof for the case of $\int_0^t f'(X_s) b(s)ds$ is identical.
