Implications of almost sure summability of stochastic processes $\newcommand{\N}{\mathbb{N}}\newcommand{\P}{\mathrm{P}}\newcommand{\esssup}{\mathrm{esssup}}$Let $(Z_j)_{j\in\N}$ be a nonnegative stochastic process ($Z_j\geq 0$ $\P$-a.s) for which the following holds:
$$
\esssup\left(\sum_{j=0}^{\infty}Z_j\right)\leq M\tag{1}\label{1},
$$
for some $M\geq 0$.
Using the definition of essential supremum we have that the sequence is almost surely summable:
$$
\begin{align}
&\P\left[ \{\omega\in\Omega: \sum_{j=0}^{\infty}Z_j > M\}\right] = 0
\\
\Leftrightarrow\ &\P\left[ \{\omega\in\Omega: \sum_{j=0}^{\infty}Z_j \leq M\}\right] = 1
\end{align}$$
and since $\{\omega\in\Omega: \sum_{j=0}^{\infty}Z_j(\omega) \leq M\}\subseteq \{\omega\in\Omega: Z_j(\omega)\to 0\}$, we have that \eqref{1} implies $\P[\{\omega: Z_j(\omega)\to 0\}]=1$, that is, $Z_j\to 0$ a.s., and, a fortiori, $Z_j\to 0$ in probability.
I wonder whether we can prove that $\esssup(Z_j)\to 0$. 
If not, there will be a $\delta>0$ and an increasing sequence of integers $(j_k)_k$ so that $\esssup(Z_{j_k})>\delta$ for all $k\in\N$. Since $Z_j$ are nonnegative, I guess (but I am not sure) that we can show that $\esssup(\sum_{k\in\N}Z_{j_k})>\delta$ which is a contradiction. Is this correct? If not, is there a counterexample where \eqref{1} holds but $\esssup(Z_j)$ does not converge to $0$?
 A: No, this need not be the case. For instance, it is possible that with probability $1$, there exists $j$ such that $Z_j=M$ and $Z_k=0$ for all $k\neq j$. For an explicit example, let $X$ be a Poisson random variable and define $Z_j:=\mathbf1_{\{X=j\}}$ for each $j$. Then $\sum_{j=0}^\infty Z_j=1$ almost surely, but $\operatorname{esssup}(Z_j)=1$ for all $j$, so in particular it does not converge to zero.
Note that the conclusion you are trying to draw - that $Z_j\to0$ in $L^\infty$ - is extremely strong. This is rarely something we would try to prove in probability theory.
EDIT: As we have discussed in the comments, it is true that $Z_j\to0$ in $L^p$ for any $1\le p<\infty$. To see this, it is sufficient to show that there is a constant $C$ such that $\sum_jZ_j^p\le C$ almost surely, since this would imply $\sum_j\mathbb E[Z_j^p]=\mathbb E\left[\sum_jZ_j^p\right]\le C$ by the monotone convergence theorem, which of course shows that $\mathbb E[Z_j^p]\to0$.
Let $A=\{j\in\mathbb N:Z_j\le1\}$ and $B=\mathbb N\setminus A$. Notice that $Z_j^p\le Z_j$ for all $j\in A$, $Z_j\le M$ for all $j\in B$ almost surely, and $|B|\le\sum_{j\in B}Z_j\le M$. This implies
$$\sum_jZ_j^p=\sum_{j\in A}Z_j^p+\sum_{j\in B}Z_j^p\le\sum_{j\in A}Z_j+M\cdot M^p\le M+M^{p+1}$$
almost surely, completing the proof.
