I am studying the following ito lemma proof, however I am having some trouble understading, and it has been a week without figurigh out certain steps. Here is the theorem statement and full proof, scrolling down you can find the question section.
Theorem 5.1. Let $M$ be a continuous local martingale and $V$ be a continuous process which is locally of bounded variation. Let $f$ be a continuous real-valued function defined on $\mathbb{R}^{2}$ such that the partial derivatives $\frac{\partial f}{\partial x}(x, y), \frac{\partial^{2} f}{\partial x^{2}}(x, y)$, and $\frac{\partial f}{\partial y}(x, y)$, exist and are continuous for all $(x, y)$ in $\mathbb{R}^{2}$. Then a.s., we have for each $t$
$$ \begin{aligned} f\left(M_{t}, V_{t}\right)-f\left(M_{0}, V_{0}\right)= & \int_{0}^{t} \frac{\partial f}{\partial x}\left(M_{s}, V_{s}\right) d M_{s} \\ & +\int_{0}^{t} \frac{\partial f}{\partial y}\left(M_{s}, V_{s}\right) d V_{s} \\ & +\frac{1}{2} \int_{0}^{t} \frac{\partial^{2} f}{\partial x^{2}}\left(M_{s}, V_{s}\right) d[M]_{s} \end{aligned} $$
For clarity, we have put in the time-parameter $s$ in the stochastic integral $\int_{0}^{t} \frac{\partial f}{\partial x}\left(M_{s}, V_{s}\right) d M_{s}$. The reader should keep in mind that this integral is defined stochastically, not path-by-path. A suggestive way to write (5.2) is by using differentials:
$$ \begin{aligned} d f\left(M_{t}, V_{t}\right)= & \frac{\partial f}{\partial x}\left(M_{t}, V_{t}\right) d M_{t} \\ & +\frac{\partial f}{\partial y}\left(M_{t}, V_{t}\right) d V_{t}+\frac{1}{2} \frac{\partial^{2} f}{\partial x^{2}}\left(M_{t}, V_{t}\right) d[M]_{t} . \end{aligned} $$
Of course the rigorous interpretation of (5.3) is the integrated form (5.2). Example. Let $B$ denote a Brownian motion in $I R$ and let $f(x)=x^{2}$. With $M=B$ and $f(x, y)=f(x),(5.3)$ becomes
$$ >d\left(B_{t}\right)^{2}=2 B_{t} d B_{t}+d t $$
Formally this suggests $\left(d B_{t}\right)^{2}=d t$. For general $M$ the appropriate formalism is $\left(d M_{t}\right)^{2}=d[M]_{t}$. Heuristically this explains the presence of the additional term in the Itô formula.
Proof of Theorem 5.1. Since both sides of the equality in (5.2) are continuous processes, it suffices to prove for each $t$ that (5.2) holds a.s. Let $\left\{\pi_{t}^{n}, n \in I N\right\}$ be a sequence of partitions of $[0, t]$ such that $\lim _{n \rightarrow \infty} \delta \pi_{t}^{n}=0$. We use the same notation for members of $\pi_{t}^{n}$ as in Chapter 4. In particular, we omit the superscript $n$ from $t_{j}^{n}$. Thus,
$$ \begin{aligned} f\left(M_{t}, V_{t}\right)-f\left(M_{0}, V_{0}\right)=\sum_{j}\{ & f\left(M_{t_{j+1}}, V_{t_{j+1}}\right) \\ & -f\left(M_{t_{j+1}}, V_{t_{j}}\right) \\ & \left.+f\left(M_{t_{j+1}}, V_{t_{j}}\right)-f\left(M_{t_{j}}, V_{t_{j}}\right)\right\} . \end{aligned} $$
By Taylor's theorem, the right side of the equals sign above may be written as:
$$ \begin{aligned} & \sum_{j}\left\{\left(\frac{\partial f}{\partial y}\right)\left(M_{t_{j}}, V_{t_{j}}\right)+\varepsilon_{j}^{1}\right)\left(V_{t_{j+1}}-V_{t_{j}}\right) \\ &+\frac{\partial f}{\partial x}\left(M_{t_{j}}, V_{t_{j}}\right)\left(M_{t_{j+1}}-M_{t_{j}}\right) \\ &\left.+\frac{1}{2}\left(\frac{\partial^{2} f}{\partial x^{2}}\left(M_{t_{j}}, V_{t_{j}}\right)+\varepsilon_{j}^{2}\right)\left(M_{t_{j+1}}-M_{t_{j}}\right)^{2}\right\} \end{aligned} $$
where
$$ \varepsilon_{j}^{1}=\frac{\partial f}{\partial y}\left(M_{t_{j+1}}, V_{\tau_{j}}\right)-\frac{\partial f}{\partial y}\left(M_{t_{j}}, V_{t_{j}}\right) $$
and
$$ \varepsilon_{j}^{2}=\frac{\partial^{2} f}{\partial x^{2}}\left(M_{\eta_{j}}, V_{t_{j}}\right)-\frac{\partial^{2} f}{\partial x^{2}}\left(M_{t_{j}}, V_{t_{j}}\right) $$
for some random times $\tau_{j}$ and $\eta_{j}$ in $\left[t_{j}, t_{j+1}\right]$. For each $\omega$, the functions $(r, s) \rightarrow \frac{\partial f}{\partial y}\left(M_{r}, V_{s}\right)(\omega)$ and $(r, s) \rightarrow \frac{\partial^{2} f}{\partial x^{2}}\left(M_{r}, V_{s}\right)(\omega)$ are uniformly continuous on $[0, t]^{2}$, and hence $\sup _{j}\left|\varepsilon_{j}^{1}(\omega)\right|$ and $\sup _{j}\left|\varepsilon_{j}^{2}(\omega)\right|$ tend to zero as $n \rightarrow \infty$. (Note that $\varepsilon_{j}^{1}$ and $\varepsilon_{j}^{2}$ depend on $n$ although the notation does not specifically indicate this because the indices $n$ on the $t_{j}^{n}$ have been suppressed.) From this property of $\varepsilon_{j}^{1}(\omega)$, the continuity of $s \rightarrow \frac{\partial f}{\partial y}\left(M_{s}, V_{s}\right)(\omega)$, and since $s \rightarrow V_{s}(\omega)$ is of bounded variation on $[0, t]$ for almost every $\omega$, it follows that
$$ >\sum_{j}\left(\frac{\partial f}{\partial y}\left(M_{t_{j}}, V_{t_{j}}\right)+\varepsilon_{j}^{1}\right)\left(V_{t_{j+1}}-V_{t_{j}}\right) \rightarrow \int_{0}^{t} \frac{\partial f}{\partial y}\left(M_{s}, V_{s}\right) d V_{s} $$
almost surely as $n \rightarrow \infty$. From the above property of $\varepsilon_{j}^{2}(\omega)$, and since $\sum_{j}\left(M_{t_{j+1}}-M_{t_{j}}\right)^{2} \rightarrow[M]_{t}$ in pr. as $n \rightarrow \infty$ by Theorem 4.1(ii), it follows that $\sum_{j} \varepsilon_{j}^{2}\left(M_{t_{j+1}}-M_{t_{j}}\right)^{2} \rightarrow 0$ in pr. as $n \rightarrow \infty$.
The proof will be completed in two steps. First we prove that when $M$ and $V$ are bounded, the terms in (5.5) involving the $x$-partial derivatives of $f$ converge in pr. to the appropriate terms in (5.2). Then we extend (5.2) to the general case by using a localizing sequence for $M$ and $V$.
Suppose that $M$ and $V$ are bounded. Then $\frac{\partial f}{\partial x}$ and $\frac{\partial^{2} f}{\partial x^{2}}$ are bounded on the range of $(M, V)$ and $\mu_{M}$ is a finite measure on $\mathcal{P}$. For each $n$, the process $X^{n}$ defined by
$$ X^{n}=\sum_{j} \frac{\partial f}{\partial x}\left(M_{t_{j}}, V_{t_{j}}\right) 1_{\left(t_{j}, t_{j+1}\right]}+\frac{\partial f}{\partial x}\left(M_{0}, V_{0}\right) 1_{\{0\}} $$
is predictable and $\left\{X^{n}\right\}$ converges pointwise to $1_{[0, t]} \frac{\partial f}{\partial x}(M, V)$ on $\mathbb{R}_{+} \times \Omega$ and hence by bounded convergence in $\mathcal{L}^{2}$. Then by the isometry,
$$ \begin{aligned} \sum_{j} \frac{\partial f}{\partial x}\left(M_{t_{j}}, V_{t_{j}}\right)\left(M_{t_{j+1}}-M_{t_{j}}\right) & =\int X_{s}^{n} d M_{s} \\ & \rightarrow \int_{0}^{t} \frac{\partial f}{\partial x}\left(M_{s}, V_{s}\right) d M_{s} \end{aligned} $$
in $L^{2}$ as $n \rightarrow \infty$. Also, by Theorem 4.8 with $Y=\frac{\partial^{2} f}{\partial x^{2}}(M, V)$, we have
$$ >\sum_{j} \frac{\partial^{2} f}{\partial x^{2}}\left(M_{t_{j}}, V_{t_{j}}\right)\left(M_{t_{j+1}}-M_{t_{j}}\right)^{2} \rightarrow \int_{0}^{t} \frac{\partial^{2} f}{\partial x^{2}}\left(M_{s}, V_{s}\right) d[M]_{s} $$
in pr. as $n \rightarrow \infty$. It follows that the expression in (5.5) converges in pr. to the right side of (5.2) and hence (5.2) holds a.s. Thus we have proved the theorem when $M$ and $V$ are bounded.
To extend to the general case, for each $n$ let $\tau_{n}=\inf \{t \geq 0$ : $\left.\left|M_{t}\right| \vee\left|V_{t}\right|>n\right\}$. Then $M^{n}=M \cdot \wedge \tau_{n} 1_{\left\{\tau_{n}>0\right\}}$ and $V^{n}=V \cdot \wedge \tau_{n} 1_{\left\{\tau_{n}>0\right\}}$ are bounded, and it follows from the above that (5.2) holds a.s. with $M^{n}$, $V^{n}$ in place of $M, V$, respectively, and hence holds a.s. with $t \wedge \tau_{n}$ in place of $t$. Then (5.2) follows on letting $n \rightarrow \infty$.
Question
My problem starts in the following step: $$ \begin{aligned} & \sum_{j}\left\{\left(\frac{\partial f}{\partial y}\right)\left(M_{t_{j}}, V_{t_{j}}\right)+\varepsilon_{j}^{1}\right)\left(V_{t_{j+1}}-V_{t_{j}}\right) \\ &+\frac{\partial f}{\partial x}\left(M_{t_{j}}, V_{t_{j}}\right)\left(M_{t_{j+1}}-M_{t_{j}}\right) \\ &\left.+\frac{1}{2}\left(\frac{\partial^{2} f}{\partial x^{2}}\left(M_{t_{j}}, V_{t_{j}}\right)+\varepsilon_{j}^{2}\right)\left(M_{t_{j+1}}-M_{t_{j}}\right)^{2}\right\} \end{aligned} $$
where
$$ \varepsilon_{j}^{1}=\frac{\partial f}{\partial y}\left(M_{t_{j+1}}, V_{\tau_{j}}\right)-\frac{\partial f}{\partial y}\left(M_{t_{j}}, V_{t_{j}}\right) $$
and
$$ \varepsilon_{j}^{2}=\frac{\partial^{2} f}{\partial x^{2}}\left(M_{\eta_{j}}, V_{t_{j}}\right)-\frac{\partial^{2} f}{\partial x^{2}}\left(M_{t_{j}}, V_{t_{j}}\right) $$
for some random times $\tau_{j}$ and $\eta_{j}$ in $\left[t_{j}, t_{j+1}\right]$.
How is the author applying in this case the Taylor theorem? I noticed that the author is incrementing $\frac{\partial f}{\partial y}(M_{t_j},V_{t_j})$ with and the same for $\frac{\partial^2 f}{\partial x^2}(M_{t_j},V_{t_j})$ with $\epsilon_1$ and $\epsilon_2$ respectively.But why does the author use $\tau_{j}$ and $\eta_{j}$ instead of $t_j$? I think the latter would resume the standard Taylor expansion formula.
For each $\omega$, the functions $(r, s) \rightarrow \frac{\partial f}{\partial y}\left(M_{r}, V_{s}\right)(\omega)$ and $(r, s) \rightarrow \frac{\partial^{2} f}{\partial x^{2}}\left(M_{r}, V_{s}\right)(\omega)$ are uniformly continuous on $[0, t]^{2}$, and hence $\sup _{j}\left|\varepsilon_{j}^{1}(\omega)\right|$ and $\sup _{j}\left|\varepsilon_{j}^{2}(\omega)\right|$ tend to zero as $n \rightarrow \infty$.
The partial derivative of $f$ is assumed to be continuous but how does he infer the partial derivatives, hence $\epsilon_1$ and $\epsilon_2$ are uniformly continuous? That does not seem straightforward to me.
Thanks in advance!
The authors use a version of Taylor formula, called Taylor-Lagrange formula. It says, for a differentiable function $g:\mathbb{R}\to \mathbb{R}$ we have that, for $a,b$ two fixed real numbers, there exists $\xi\in (a,b)$ such that
$$g(a)=g(b)+g'(\xi)(a-b)$$ or at order $2$ $$g(a)=g(b)+g'(b)(a-b)+\frac{1}{2}g''(\xi)(a-b)^2.$$ If I apply this to $f(M_{t_{j+1}},V_{t_{j+1}})-f(M_{t_{j+1}},V_{t_{j}})$ in the second variable, I get that there exists $\xi_j\in(V_{t_{j+1}},V_{t_{j}})$ (or $(V_{t_j},V_{t_{j+1}})$ depending on how they arrange) such that $$ f(M_{t_{j+1}},V_{t_{j+1}})-f(M_{t_{j+1}},V_{t_{j}})=\frac{\partial f}{\partial_y}f(M_{t_{j+1}},\xi_j)(V_{t_{j+1}}-V_{t_j}) $$ but because $V$ is continuous, I can find rather than $\xi_j$ a time $\tau_j\in(t_j,t_{j+1})$ such that $\xi_j=V_{\tau_j}$. Of course, since $V$ is random, so do $\tau_j$. This, gives $$ f(M_{t_{j+1}},V_{t_{j+1}})-f(M_{t_{j+1}},V_{t_{j}})=\frac{\partial f}{\partial_y}f(M_{t_{j+1}},V_{\tau_j})(V_{t_{j+1}}-V_{t_j}). $$ For the part concerning, we need the order $2$, because $M$ is not of finite variation but of finite quadratic variation.
Concerning your second question, the function, for almost all $\omega$, $$ (s,r)\to\frac{\partial f}{\partial_y}(M_{s},V_{r})(\omega) $$ is continuous on the compact $[0,t]$, thus it is uniformly continuous (Heine Thm). Continuity, gives that, because $V_{t_j}<V_{\tau_j}<V_{t_{j+1}}$ (or $ V_{t_{j+1}}<V_{\tau_j}<V_{t_{j}})$, for almost all $\omega$ $$ \varepsilon_j^1=\frac{\partial f}{\partial y}\left(M_{t_{j+1}}, V_{\tau_{j}}\right)(\omega)-\frac{\partial f}{\partial y}\left(M_{t_{j}}, V_{t_{j}}\right)(\omega)\xrightarrow[n\to\infty]{}0, $$ but for any $j$ $$ \left|\frac{\partial f}{\partial y}\left(M_{t_{j+1}}, V_{\tau_{j}}\right)(\omega)-\frac{\partial f}{\partial y}\left(M_{t_{j}}, V_{t_{j}}\right)(\omega)\right|\leq \sup_{\underset{\scriptstyle|r-v|<\delta_n}{|s-u|<\delta_n}} \left|\frac{\partial f}{\partial y}\left(M_{s}, V_{r}\right)(\omega)-\frac{\partial f}{\partial y}\left(M_{u}, V_{v}\right)(\omega)\right|, $$ where $\delta_n=\sup_j|t_{j+1}-t_j|$. So $$ \sup_{j}\left|\frac{\partial f}{\partial y}\left(M_{t_{j+1}}, V_{\tau_{j}}\right)(\omega)-\frac{\partial f}{\partial y}\left(M_{t_{j}}, V_{t_{j}}\right)(\omega)\right|\leq \sup_{\underset{\scriptstyle|r-v|<\delta_n}{|s-u|<\delta_n}} \left|\frac{\partial f}{\partial y}\left(M_{s}, V_{r}\right)(\omega)-\frac{\partial f}{\partial y}\left(M_{u}, V_{v}\right)(\omega)\right| $$ but, uniform continuity gives $$ \sup_{\underset{\scriptstyle|r-v|<\delta_n}{|s-u|<\delta_n}} \left|\frac{\partial f}{\partial y}\left(M_{s}, V_{r}\right)(\omega)-\frac{\partial f}{\partial y}\left(M_{u}, V_{v}\right)(\omega)\right|\xrightarrow[n\to\infty]{}0. $$ It is similar for the $\varepsilon^2$'s.