Tail Probabilities of a martingale difference sequence I'm currently facing the problem, that I can neither prove nor find a counterexample for the following statement.

Let $q \in (1,2)$ and let $(D_n)_{n \in \mathbb{N}}$ be a martingale difference sequence on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ w.r.t. to a filtration $F = (F_n)_{n \in \mathbb{N}}$. Further assume that $(\mathbb{E}(\vert D_n \vert^q))^{1/q} =\Vert D_n \Vert_q < \infty$, i.e. $D_n \in L^q(\Omega,\mathcal{F})$ for all $n \in \mathbb{N}$. Then there exists a random variable $D \in L^q(\Omega,\mathcal{F})$ and a $C > 0$ such that $$\mathbb{P}(\vert D_n \vert > z) \leq C \mathbb{P}(\vert D \vert > z)$$ for all $z > 0$ and $n \in \mathbb{N}$.

My first idea was to set $D := \sup_n \vert D_n \vert$ and then apply a maximal inequality. But inequalities of this type (for example Doob's martingale inequality) seem not to work here, since we don't take the supremum over a bounded set. (And I also guess $\sup_n \vert D_n \vert$ lies not necessarily in $L^q$)
Does someone have some ideas in mind?
Although I'm not sure whether or not this is of interest: I need such a result to apply the Marcinkiewicz-Zygmund Strong Law of Large Numbers (see for example "On the convergence of weighted sums of martingale differences, by Nguyen&Nguyen Corollary 3, p. 303).
 A: This is false. A condition like
$$\exists C \in \mathbf{R}_+,~\exists D \in L^q(\mathbf{P}),~\forall n \in \mathbf{N},~\forall z \in \mathbf{R}_+^*, \mathbf{P}[|D_n|>z] \le C \mathbf{P}[|D|>z]$$
would force $(D_n)_{n \in \mathbf{N}}$ to be bounded in $L^q(\mathbf{P})$ since
$$\mathbf{E}[|D_n|^q] = \int_0^\infty \mathbf{P}[|D_n|>z]~qz^{q-1} \mathrm{d}z \le C\int_0^\infty \mathbf{P}[|D|>z]~qz^{q-1} \mathrm{d}z = C\mathbf{E}[|D|^q] .$$ This is not always the case, for ewample when the random variables $D_n/n$ are i.i.d. and not almost surely null.
I complete the answer below: assuming that $(D_n)_{n \in \mathbf{N}}$ to be bounded in $L^q(\mathbf{P})$ is not sufficient. Actually, we have the implications
\begin{eqnarray}
\text{ for some }p>q, (D_n)_{n \in \mathbf{N}} \text{ bounded in } L^q(\mathbf{P})
&\implies& \exists C \in \mathbf{R}_+,~\exists D \in L^q(\mathbf{P}),~\forall n \in \mathbf{N},~\forall z \in \mathbf{R}_+^*, \mathbf{P}[|D_n|>z] \le C \mathbf{P}[|D|>z] \\ 
&\implies& (|D_n|^q)_{n \in \mathbf{N}} \text{ uniformly integrable } 
\\
&\implies& (D_n)_{n \in \mathbf{N}} \text{ bounded in } L^q(\mathbf{P})
\end{eqnarray}
Proof of the first implication
If for some $p>q$, $(D_n)_{n \in \mathbf{N}}$ is bounded in $L^q(\mathbf{P})$, then $C:= \sup\{\mathbf{E}[|D_n|^p] : n \in \mathbf{N}\}$ is finite. For every $z \in \mathbf{R_+}^*$,
$$\mathbf{P}[|D_n|>z] \le \min(1,|z|^{-q})\mathbf{E}[|D_n|^p]$$
so
$$\mathbf{P}[|D_n|>z] \le \min(1,C|z|^{-q})$$
Let $D$ any positive random variable such that for every $z \in \mathbf{R_+}^*$, $\mathbf{P}[D>z] \le \min(1,C|z|^{-q})$. Then
$D \in L^q(\mathbf{P})$ since
$$\mathbf{E}[D^q] = \int_0^{+\infty} \mathbf{P}[D>z]~qz^{q-1} \mathrm{d}z \le \int_0^{+\infty} \min(1,C|z|^{-q})~qz^{q-1}~\mathrm{d}z < +\infty.$$
Proof of the second implication
Assume that we have $C \in \mathbf{R}_+$ and $D \in L^q(\mathbf{P})$ such that
$$\forall n \in \mathbf{N},~\forall z \in \mathbf{R}_+^*, \mathbf{P}[|D_n|>z] \le C \mathbf{P}[|D|>z].$$
Then for every $x \in \mathbf{R_+}^*$
\begin{eqnarray}
\mathbf{E}[|D_n|^q \mathbf{1}_{[|D_n|^q>x]}] 
&=& \mathbf{E}\Big[\int_0^\infty \mathbf{1}_{[|D_n|^q>x]}] \mathbf{1}_{[|D_n|>z]}~qz^{q-1} \mathrm{d}z \Big] \\
&=& \mathbf{E}\Big[\int_0^\infty \mathbf{1}_{[|D_n|>\max(x^{1/q},z)]}~qz^{q-1} \mathrm{d}z \Big] \\
&=& \int_0^\infty \mathbf{P}[|D_n|>\max(x^{1/q},z)]~qz^{q-1} \mathrm{d}z \\
&=& C \int_0^\infty \mathbf{P}[|D|>\max(x^{1/q},z)]~qz^{q-1} \mathrm{d}z \\
&\le& C\int_0^\infty \mathbf{P}[|D|>z]~qz^{q-1} \mathrm{d}z \\
&=& C\mathbf{E}[|D|^q\mathbf{1}_{[|D|^q>x]}]. 
\end{eqnarray}
By Lebesgue dominated convergence theorem, the right-hand side goes to $0$ as $x$ goes to infinity. Hence $(|D_n|^q)_{n \in \mathbf{N}}$ is  uniformly integrable.
The third implication is classical. https://en.wikipedia.org/wiki/Uniform_integrability
