Confusion about the difference of quadratic variations and the Lebesgue-Stieltjes Integral Consider a path of finite quadratic variation $\omega$ on some fixed time horizon $[0,T]$ and $f$ a function of paths on $\mathbb R$
let $[ \omega ](t)$ be the quadratic variation of the path $\omega$ at $t\in [0,T]$. I recently saw a note, stating that 
$\int f d[\omega]=\lim\limits_{n \to \infty}\sum\limits_{t_{i} \in \pi^{n}}f(\omega(t_{i}))(\omega(t_{i+1})-\omega(t_{i}))^{2}$
where $\int f d[\omega]$ is in actual fact the Lebesgue-Stieltjes Integral namely: $\int f d[\omega]=\lim\limits_{n \to \infty}\sum\limits_{t_{i} \in \pi^{n}}f(\omega(t_{i}))([\omega](t_{i+1})-[\omega](t_{i}))$
My question is why are then the following equal: 
$$\lim\limits_{n \to \infty}\sum\limits_{t_{i} \in \pi^{n}}f(\omega(t_{i}))(\omega(t_{i+1})-\omega(t_{i}))^{2}=\lim\limits_{n \to \infty}\sum\limits_{t_{i} \in \pi^{n}}f(\omega(t_{i}))([\omega](t_{i+1})-[\omega](t_{i}))$$
source of my confusion: how can 
$$(\omega(t_{i+1})-\omega(t_{i}))^{2}=\omega(t_{i+1})^{2}-2\omega(t_{i+1})\omega(t_{i})+\omega(t_{i})^{2} $$
and 
$$([\omega](t_{i+1})-[\omega](t_{i}))$$
be equal (even in the limit)?
 A: This is just my try on the question given that it hasn't received any answers yet. 
I will assume that $f$, $\omega$ and $[\omega]$ are bounded otherwise you can use stopping times such that this assumption holds and then make passage to infinity.
$$E\left[\left(\sum_{i=1}^m f(\omega(t_{i}))\{(\omega(t_{i+1})-\omega(t_{i}))^2-([\omega](t_{i+1})-[\omega](t_{i}))\}\right)^2\right]$$
$$\leq  E\left(\sum_{i=1}^m f(\omega(t_{i}))^2\{(\omega(t_{i+1})-\omega(t_{i}))^2-([\omega](t_{i+1})-[\omega](t_{i}))\}^2\right)$$
$$\leq 2\|f\|_{\infty}^2 E\left(\sum_{i=1}^m (\omega(t_{i+1})-\omega(t_{i}))^4+\sum_{i=1}^m([\omega](t_{i+1})-[\omega](t_{i}))^2\right)$$
$\sum_{i=1}^m (\omega(t_{i+1})-\omega(t_{i}))^4$ tends to $0$ when the mesh of the partition goes to $0$, see for example Lemma $1.5.10$ on Karatzas' book.
The second term is (letting the mesh go to $0$) the quadratic variation of the quadratic variation, if $\omega$ is continuous, then the latter is $0$. (see the following Quadratic Variation of Quadratic Variation)
(of course in order to take limits inside the expectation I am using some bounded convergence result)
Then you have that the expressions converge in quadratic mean.
Hopefully you'll find this useful even though it's not a proper answer; if you notice something off about it please let me know!
EDIT: As pointed out by @TheoreticalEconomist the application of the Cauchy-Schwarz was not suitable for this particual case. Instead notice that the cross-products are neglected in the first step. For an explanation on this particular see theorem $1.5.8$ on Karatzas' book, and the following discussion. 
