Writing a martingale as the difference of two non-negative martingales Assumption: $ (M_{t})_{t \geq 0}$ is a martingale w.r.t $(\mathcal{F}_t)_{t \geq 0}$
Question: Why can I write $M_t$ as the difference of two non-negative martingales?
Attempt: $ M_t = M_t^{+} - M_t^{-}$ But in general, $M_t^{+}$ is a supermartingale, and $M_t^{-}$ is a submartingale.
Can someone help me understand how to write the decomposition properly?
 A: 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.
