I know that if a function $f:[0,1]\to \mathbb R$ has finite bounded variation, then its quadratic variation is zero.
I say that $X$ has bounded variation in $L^1([0,1])$ if for all partition $\pi_n: 0=t_0<t_1<...<t_n=1$ s.t. $\delta _n:=\max|t_{i+1}-t_i|\to 0$ when $n\to \infty $, $$\sum_{i=1}^n |X_{t_{i+1}}-X_{t_i}|$$ is a Cauchy Cauchy sequence.
Suppose that $X$ is a a.s. continuous process with finite bounded variation. I was wondering if $$L^1-\lim_{n\to \infty }\sum_{i=0}^{n-1}|X_{t_{i+1}}-X_{t_i}|^2=0.$$
Attempts
I have that $$\mathbb E\left[\sum_{i=0}^{n-1}|X_{t_{i+1}}-X_{t_i}|^2\right]\leq \mathbb E\left[\max_{i=1,...,n}|X_{t_{i+1}}-X_{t_i}|\cdot \sum_{i=0}^{n-1}|X_{t_{i+1}}-X_{t_i}|\right].\tag{E}$$
So, if $X_t=\int_0^t f(s,\cdot )$ for some a.s. bounded function $f$, then $(E)$ can be upper bounded $$C\delta _n\mathbb E\left[\sum_{i=1}^n |X_{t_{i+1}}-X_{t_i}|\right]\underset{n\to \infty }{\longrightarrow }0.$$
It also work if there is $Y\in L^1$ s.t. $|f(t,\omega )|\leq Y(\omega )$ for all $t\in [0,1]$, but is it still true in a more general case ?
I'm asking this question because I want to prove that if $$X_t=\underbrace{\int_0^tf(s,\cdot )ds}_{=:A_t}+\underbrace{\int_0^t g(s,\cdot )dW_s}_{=:M_t},$$
then $$[X]_t=[M]_t=\int_0^tg(s,\cdot )^2ds.$$ And at some point, I must compute $$\mathbb E\left[\sum_{i}(A_{i+1}-A_i)(M_{i+1}-M_i)\right]\leq\sqrt{\mathbb E\sum_{i}(A_{i+1}-A_i)^2}\sqrt{\mathbb E\sum_{i}(M_{i+1}-M_i)^2}.$$ So, I would like to have $\mathbb E\sum_{i}(A_{i+1}-A_i)^2\to 0$.