Convergence of sum of triangular array of random variables Let $\left(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\geq0},\mathbb{P}\right)$ be a filtered probability space. For given $t>0$ consider the equi-spaced partition $t_{j,n}=j\,\Delta_n$ with $\Delta_n=t/n$ of $[0,t]$. In this framework, consider a triangular array of random variable $\{\xi_{j,n}\}_{j=1,\dots,n}$ such that $\xi_{j,n}\in\mathcal{F}_{j\,\Delta_n}$ for all $j$. In what follows, I denote with $\stackrel{u.c.p}{\longrightarrow}$ the uniform, on $[0,t]$, convergence in probability and with $\mathbb{E}_{j}[\cdot]$ the $\mathcal{F}_{j\,\Delta_n}$-conditional expectation.
This is the lemma I want to prove
Lemma.
If the following conditions are satisfied:
$$
\sum^{\lfloor t/\Delta_n \rfloor}_{j=1}\mathbb{E}_{j-1}[\xi_{j,n}] \stackrel{u.c.p}{\longrightarrow}0,\quad (1)
$$
$$
\sum^{\lfloor t/\Delta_n \rfloor}_{j=1}\mathbb{E}_{j-1}[\xi^2_{j,n}] \stackrel{p}{\longrightarrow}0\quad(2)
$$
then
$$
\sum_{j=1}^{\lfloor t/\Delta_n\rfloor}\xi_{j,n}\stackrel{u.c.p}{\longrightarrow}0
$$
Attempt of Proof.
For given $t\geq 0$ call $n=\lfloor t/\Delta_n \rfloor$ where $\Delta_n\longrightarrow0$ as $n\longrightarrow\infty$. We have
$$
\left|\sum_{j=1}^n\xi_{j,n}\right|\leq \sum_{j=1}^n\left|\xi_{j,n}\right| \leq \underbrace{\sum_{j=1}^n|\xi_{j,n}-\mathbb{E}_{j-1}[\xi_{j,n}]|}_{\mathsf{A}_n}+\underbrace{\sum_{j=1}^n|\mathbb{E}_{j-1}[\xi_{j,n}]|}_{\mathsf{B}_n}.
$$
For $n$ given, the discrete-time process $M_{k,n}=\sum_{j=1}^k(\xi_{j,n}-\mathbb{E}_{j-1}[\xi_{j,n}])$, with $k=1,\dots,n$, is a martingale with respect to $\mathcal{G}_{k,n}=\mathcal{F}_{k\,\Delta_n}$. In fact we have
$$
\mathbb{E}_k[M_{k+1,n}]=\mathbb{E}_{k}[M_{k,n}+\xi_{k+1,n}-\mathbb{E}_{k}[\xi_{k+1,n}]] = M_{k,n},
$$
where the $\mathcal{G}_{k,n}$-measurability of $M_{k,n}$ derives from the
the $\mathcal{F}_{k\,\Delta_n}$-measurability of $\xi_{k,n}$. The idea is that condition (1) should imply $\mathsf{B}_n\stackrel{p}{\longrightarrow}0$ and the Doob's martingale inequality applied to $M_{k,n}$ plus condition (2) should imply $\mathsf{A}_n\stackrel{p}{\longrightarrow}0$. The problems that I face are thus:
I) Condition (1) is equivalent to
$$
\sup_{s\in[0,t]}\left|\sum^{\lfloor s/\Delta_n \rfloor}_{j=1}\mathbb{E}_{j-1}[\xi_{j,n}]\right|\stackrel{p}{\rightarrow}0
$$
which is weaker (right?) than $\sum_{j=1}^n|\mathbb{E}_{j-1}[\xi_{j,n}]|\stackrel{p}{\longrightarrow}0$.
II) The  Doob's martingale inequality (with $p=2$) prescribes that
$$
\mathbb{E}\left[\left(\max_{q\leq k}M_{q,n}\right)^2\right]\leq C\,\mathbb{E}\left[M^2_{k,n}\right]
$$
A simple computation shows that
$$
\mathbb{E}\left[M^2_{k,n}\right] = \sum_{j=1}^k\mathbb{E}\left[\xi_{j,n}^2-\mathbb{E}_{j-1}\left[\xi_{j,n}\right]^2\right]
$$
and, as for point I), I don't know how to use condition (1) or (2) to finish the proof.
Am I missing something obvious? Or, perhaps, am I moving on a dead end track?
 A: In your first line, plus-and-subtract before using triangle inequality,
$$\biggl|\sum_{j=1}^{n}\xi_{j,n}\biggr|=\biggl|\sum_{j=1}^{n}(\xi_{j,n}-\mathbb{E}_{j-1}[\xi_{j,n}])+\sum_{j=1}^{n}\mathbb{E}_{j-1}[\xi_{j,n}]\biggr|\leq\biggl|\sum_{j=1}^{n}(\xi_{j,n}-\mathbb{E}_{j-1}[\xi_{j,n}])\biggr|+\biggl|\sum_{j=1}^{n}\mathbb{E}_{j-1}[\xi_{j,n}]\biggr|.$$
It gives a stronger upper bound: above line only uses the triangle inequality once (counting of the basic form $|a+b|\leq|a|+|b|$), while your attempt uses $2(n-1)$ times.
A: The following will give a proof of what you want. Using your notations, let
\begin{align*}
 M_{k,n}&=\sum_{j=1}^{k}(\xi_{j,n}- \mathsf{E}_{j-1}[\xi_{j,n}]),
 \quad k\le l =\Big[\frac{t}{\Delta_n}\Big],  \\
   \langle M\rangle_{k,n}&=\sum_{j=1}^{k}\mathsf{E}_{j-1}[ (\xi_{j,n}- \mathsf{E}_{j-1}[\xi_{j,n}])^2 ], \qquad k\le l.
\end{align*}
Then it is easy to verify that $\{M_{k,n}^2-\langle M\rangle_{k,n},\; 1\le k \le l\} $
is a martingale for each $n\ge 1$ and following inequality is true for $\epsilon>0, \delta>0$,
\begin{equation*}
 \mathsf{P}(\sup_{k\le l}|M_{k,n}|\ge \delta)\le \frac{\epsilon}{\delta^2} 
 +\mathsf{P}(\langle M\rangle_{l,n}\ge \epsilon). \tag{3}
\end{equation*}
(The proof of (3) and its more general cases, please refer to
J. Jacod, and  A. N. Shiryayev, Limit Theory for Stochastic Processes, 2ed. Springer, 2003.
p.35, Lemma 1.3.30. ).
Meanwhile,
\begin{gather*}
 (2): \sum_{j=1}^{l}\mathsf{E}_{j-1}[\xi_{j,n}^2]\overset{p}{\longrightarrow}0\\
 \text{$\;$}\hspace{3cm}\Downarrow  (\text{by } \mathsf{E}_{j-1}[(\xi_{j,n} - \mathsf{E}_{j-1}[\xi_{j,n}])^2] 
    \le  \mathsf{E}_{j-1}[\xi_{j,n}^2])  \\
 \langle M\rangle_{l,n} 
    =\sum_{j=1}^{l}\mathsf{E}_{j-1}[(\xi_{j,n} - \mathsf{E}_{j-1}[\xi_{j,n}])^2] \overset{p}{\longrightarrow}0\\
 \Downarrow (\text{by } (3))\\
 \sup_{k\le l}\Big|\sum_{j=1}^{k}(\xi_{j,n}- 
    \mathsf{E}_{j-1}[\xi_{j,n}])\Big| =
 \sup_{k\le l}|M_{k,n}| \overset{p}{\longrightarrow}0\\
 \Downarrow (\text{by } (1))\\
 \sup_{k\le l}\Big|\sum_{j=1}^{k}\xi_{j,n}\Big|\overset{p}{\longrightarrow}0.
\end{gather*}
