First of all, note that we need a further assumption on $(M_t)_{t \geq 0}$; namely that
$$\lim_{t \to \infty} \mathbb{E}|M_t|<\infty. \tag{1}$$
It can be easily seen that if there exists a decomposition of the form $M_t = U_t-V_t$ where $(U_t)_{t \geq 0}$ and $(V_t)_{t \geq 0}$ are non-negative martingales with respect to the filtration $(\mathcal{F}_t)_{t \geq 0}$, then $(1)$ holds. Indeed,
$$\mathbb{E}|M_t| \leq \mathbb{E}|U_t| + \mathbb{E}|V_t| = \mathbb{E}U_t+\mathbb{E}V_t = \mathbb{E}U_0+\mathbb{E}V_0$$
where we used in the last step that both $(U_t)_{t \geq 0}$ and $(V_t)_{t \geq 0}$ are martingales.
On the other hand, $(1)$ implies the existence of such a decomposition as the following proof shows: Since $x \mapsto x^+$ and $x \mapsto x^-$ are convex functions, we know from Jensen's inequality that $(M_t^+)_{t \geq 0}$ and $(M_t^-)_{t \geq 0}$ are submartingales. In particular,
$$\mathbb{E}(M_{t+s}^+ \mid \mathcal{F}_t) \geq \mathbb{E}(M_{t+r}^+ \mid \mathcal{F}_t)$$
for any $0 \leq r \leq s$. Therefore, it follows from the monotone convergence theorem and $(1)$ that
$$U_t := \lim_{s \to \infty} \mathbb{E}(M_{t+s}^+ \mid \mathcal{F}_t) \geq 0$$
exists. Moreover, by the tower property, $(U_t)_{t \geq 0}$ is a martingale. Similarly,
$$V_t := \lim_{s \to \infty} \mathbb{E}(M_{t+s}^- \mid \mathcal{F}_t) \geq 0$$
defines a non-negative martingale. Finally, we have
$$\begin{align*} U_t-V_t &= \lim_{s \to \infty} \mathbb{E}(M_{t+s}^+ - M_{t+s}^- \mid \mathcal{F}_t) \\ &= \lim_{s \to \infty} \mathbb{E}(M_{t+s} \mid \mathcal{F}_t) \\ &= M_t.\end{align*}$$
This decomposition is known as Krickeberg decomposition.
Let $(\Omega,\mathcal{A},\mathbb{P})$ be a probability space.
Example 1: For any filtration $(\mathcal{F}_n)_{n \geq 1}$ and any $X \in L^1$ the process $M_n := \mathbb{E}(X \mid \mathcal{F}_n)$ is a martingale.
Example 2: Let $(\mathcal{F}_n)_{n \geq 1}$ be a filtration and $\mu$ a finite measure on $\mathcal{F}_{\infty} := \sigma(\mathcal{F}_n; n \geq 1)$. Assume that $\mu|_{\mathcal{F}_n}$ is absolutely continuous with respect to $\mathbb{P}|_{\mathcal{F}_n}$, and denote by $M_n$ the Radon-Nikodym density. Then $(M_n)_{n \geq 1}$ is a martingale.
Example 3: Consider $\Omega := (0,1)$ endowed with the Lebesgue measure, and let $(a_n)_{n \in \mathbb{N}} \subseteq (0,1)$ be a sequence of monotonically decreasing numbers. Then $$M_n := \frac{1}{a_n} 1_{(0,a_n)}$$ is a martingale.
Best Answer
There is the following result:
Note that b) requires weaker assumption on the process (martingale property is replaced by submartingale property) but the price which we have to pay is that we need to assume additionally that $F$ is increasing.
Since the modulus $F(x)=|x|$ is clearly not increasing, this means that $(X_t)_{t \geq 0}$ being a submartingale does not imply that $(|X_t|)_{t \geq 0}$ is a submartingale. An example is for instance the deterministic submartingale $X_t := - \frac{1}{1+t^2}$.