Let $X:\Omega\times[0,\infty)\to\mathbb R$ be a submartingale on a filtered probability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge0},\operatorname P)$ with $X_0=0$.
How can we show that$^1$ $\operatorname E\left[V_\infty(X)\right]<\infty$?
Clearly, if $\varsigma=\left(t_0,\ldots,t_k\right)\in\mathcal P_{[0,\:t]}$, then $$\operatorname E\left[X_{t_i}\mid\mathcal F_{t_{i-1}}\right]\ge X_{t_{i-1}}\tag1$$ and hence $$\operatorname E\left[X_{t_i}-X_{t_{i-1}}\right]\ge0\tag2$$ for all $i\in\left\{1,\ldots,k\right\}$. But how can we obtain the claim from this observation?
$^1$ Let $$\mathcal P_{[0,\:t]}:=\left\{\left(t_0,\ldots,t_k\right):k\in\mathbb N\text{ and }0=t_0<\cdots<t_k=t\right\}$$ and $$\left|\varsigma\right|:=\max_{1\le i\le k}\left(t_i-t_{i-1}\right)\;\;\;\text{for }\varsigma=\left(t_0,\ldots,t_k\right)\in\mathcal P_{[0,\:t]}\;.$$ for $t>0$. Moreover, let $$V_\varsigma(f):=\sum_{i=1}^k\left|f(t_i)-f(t_{i-1})\right|\;\;\;\text{for }\varsigma=\left(t_0,\ldots,t_k\right)\in\mathcal P_{[0,\:t]}$$ and $$V_t(f):=\sup_{\varsigma\:\in\:\mathcal P_{[0,\:t]}}V_\varsigma(f)$$ for $f:[0,t]\to\mathbb R$ and $t>0$. Now, let $$V_\infty(f):=\sup_{t>0}V_t(f)\;.$$
Example: If $X$ is standard 1-dimensional Brownian motion (a submartingale), and if $t>0$, and if $\varsigma$ is the partition of $[0,t]$ into $k$ intervals of equal length, then $$ {\operatorname E}[V_\varsigma(X)]=k\cdot \sqrt{2/k\pi}=\sqrt{2k/\pi}. $$ It follows that ${\operatorname E}[V_t(X)]=+\infty$ and also ${\operatorname E}[V_\infty(X)]=+\infty$.