I am trying to prove the following statement but I am totally at a loss.
Let $(A_t)$, $t \in \mathbb{R}^+$ be an adapted (with respect to the filtration $(\mathcal{F}_t)$) continuous integrable nondecreasing process. Let $T > 0$ and $\Pi = \{0 = t_0 < t_1 <\ldots <t_k = T\}$ be a finite partition of $[0,T]$. Define $$A_T^{\Pi} = \sum_{i=0}^{k-1}E[A_{t_{i+1}}-A_{t_i}\mid\mathcal{F}_{t_i}]$$ Then $A_T^{\Pi} \rightarrow A_T$ in $L^1$ as $\text{mesh}(\Pi)\rightarrow 0$.
There is a hint that suggests that one first attempt the case for which $E[A_T^2]<\infty$ and show that the convergence above is in $L^2$. I don't even know where to begin as I have no clue what is so special about nondecreasing continuous processes that could make this statement true. I would appreciate any hint telling me where to start even.
Now I need this result to prove the following second statement. The meaning of the subscripts is the same as in the statement above.
Let $M$ be a continuous $L^2$ martingale. Show that $$\sum_{i=0}^{k-1}E[(M_{t_{i+1}}-M_{t_i})^2\mid \mathcal{F}_{t_i}]\rightarrow [M]_T \quad \text{in} \quad L^1$$ as $\text{mesh}(\Pi)\rightarrow 0$. The question explicitly asks us to make use of the first statement.
Here is my attempt. From the first statement I have that $$\sum_{i=0}^{k-1}E[[M]_{t_{i+1}}-[M]_{t_i}\mid\mathcal{F}_{t_i}]\rightarrow [M]_T\quad \text{in} \quad L^1$$ Now for each interval $[t_i,t_{i+1}]$ I choose a submesh $\Pi_i = \{t_i = s_{i,0} < s_{i,1} <\ldots<s_{i,n_i}=t_{i+1} \}$. Since $M$ is an $L^2$-martingale, it holds that $$\sum_{j=0}^{n_i-1}(M_{s_{i,j+1}}-M_{s_{i,j}})^2 \rightarrow [M]_{t_{i+1}}-[M]_{t_{i}}\quad\text{in}\quad L^1$$ This implies $$\sum_{j=0}^{n_i-1}E[(M_{s_{i,j+1}}-M_{s_{i,j}})^2\mid\mathcal{F}_{t_i}]\rightarrow E[[M]_{t_{i+1}}-[M]_{t_{i}}\mid\mathcal{F}_{t_i}]\quad\text{in}\quad L^1$$
To see this why this implication is true note the following fact. Let $X_n \rightarrow Y$ in $L^1$ where $(X_n)$ and $Y$ are random variables defined on some probability space $(\Omega,\mathcal{F},P)$. Take an arbitrary $\sigma$-algebra $\mathcal{G}\subseteq\mathcal{F}$. Then $$E[\lvert E[X_n\mid\mathcal{G}] - E[Y\mid\mathcal{G}]\rvert] \leq E[E[\lvert X_n-Y\rvert\mid\mathcal{G}]] = E[\lvert X_n-Y\rvert]$$ The inequality above comes from Jensen. This shows that $E[X_n\mid\mathcal{G}] \rightarrow E[Y\mid\mathcal{G}]$ in $L^1$.
Now back to the question itself. With a simple trick I show that $$\sum_{j=0}^{n_i-1}E[(M_{s_{i,j+1}}-M_{s_{i,j}})^2\mid\mathcal{F}_{t_i}] = E[M^2_{t_{i+1}}-M^2_{t_i}\mid\mathcal{F}_{t_i}]$$ I also have that $$E[M^2_{t_{i+1}}-M^2_{t_i}\mid\mathcal{F}_{t_i}]=E[(M_{t_{i+1}}-M_{t_i})^2\mid\mathcal{F}_{t_i}]$$ So then $$E[(M_{t_{i+1}}-M_{t_i})^2\mid\mathcal{F}_{t_i}]\rightarrow E[[M]_{t_{i+1}}-[M]_{t_{i}}\mid\mathcal{F}_{t_i}]\quad\text{in}\quad L^1$$ This is saying that $$E[(M_{t_{i+1}}-M_{t_i})^2\mid\mathcal{F}_{t_i}]=E[[M]_{t_{i+1}}-[M]_{t_{i}}\mid\mathcal{F}_{t_i}]$$ Summing over $i$ gives $$\sum_{i=0}^{k-1}E[(M_{t_{i+1}}-M_{t_i})^2\mid \mathcal{F}_{t_i}] = \sum_{i=0}^{k-1}E[[M]_{t_{i+1}}-[M]_{t_i}\mid \mathcal{F}_{t_i}]$$ The RHS goes to $[M]_T$ in $L^1$ due to the first statement which I haven't been able to prove.
So then I have two questions:
1) How should I go about showing the first statement?
2) Is my proof for the second statement correct?