proof of Martingale criterion

Let ${({\tau }_k)}_{k\ge 1}$ be a localizing sequence of stopping times for $X$. Then:

${\mathrm{\Lambda }}_{\{{\tau }_k\ge n\}}$$\uparrow {\mathrm{\Lambda }}_{\mathrm{\Omega }}\text{a.s.}k\to \mathrm{\infty },\forall n\in \mathbb{N}$

since $\{{\tau }_k\ge n\}{\uparrow }_k{\bigcup }_1^{\mathrm{\infty }}\{{\tau }_k\ge n\}\forall n\in \mathbb{N}$.

${\displaystyle \bigcup _{k=1}^{\mathrm{\infty }}}\{{\tau }_k\ge n\}$$=\mathrm{\Omega }\text{a.s., since}{\tau }_k\to \mathrm{\infty },\text{a.s.}$

Now assume (the case being analogous).

1) We have .

We proceed by (backward) induction. For $n=n_0$ the statement holds.

$n\mapsto n-1$:

${(X^{{\tau }_k})}^{-}$$={(X_{{\tau }_k\wedge n}^{-})}_{n\in \mathbb{N}}\text{submartingale}$

We have:

${\displaystyle {\int }_{\{{\tau }_k\ge n\}}}{X}_{n-1}^{-}𝑑P$$={\displaystyle {\int }_{\{{\tau }_k\ge n\}}}{X}_{{\tau }_k\wedge (n-1)}^{-}𝑑P$

$\le {\displaystyle {\int }_{\{{\tau }_k\ge n\}}}{X}_{{\tau }_k\wedge n}^{-}𝑑P={\displaystyle {\int }_{\{{\tau }_k\ge n\}}}{X}_n^{-}𝑑P$

Where the first to second line is the submartingale property and the last line follows by induction hypothesis.

Using Fatou we get:

${\displaystyle \int X_{n-1}^{-}𝑑P}$$={\displaystyle \int \underset{k\to \mathrm{\infty }}{lim}{X}_{n-1}^{-}{\mathrm{\Lambda }}_{\{{\tau }_k\ge n\}}dP}$

$\le \underset{k\to \mathrm{\infty }}{lim\; inf}{\displaystyle \int X_{n-1}^{-}{\mathrm{\Lambda }}_{\{{\tau }_k\ge n\}}𝑑P}$

2) We have $X_n\in {ℒ}^1(n\in \mathbb{N})$.

We have $X_{{\tau }_k\wedge n}^{+}\to X_n^{+}$ a.s., $k\to \mathrm{\infty },\forall n\in \mathbb{N}$. With Fatou we get:

$E{X}_n^{+}$$\le \underset{k\to \mathrm{\infty }}{lim\; inf}E{X}_{{\tau }_k\wedge n}^{+}$

$=E{X}_0+\underset{k\to \mathrm{\infty }}{lim\; inf}E{X}_{{\tau }_k\wedge n}^{-}$

$=E{X}_0+\underset{k\to \mathrm{\infty }}{lim\; inf}E\left({\displaystyle \sum _{j=0}^{n-1}}{X}_j^{-}{\mathrm{\Lambda }}_{\{{\tau }_k=j\}}+X_n^{-}{\mathrm{\Lambda }}_{\{{\tau }_k\ge n\}}\right)$

With 1) $X_n\in {ℒ}^1$ follows.

3)

$X$ is a martingale, because $X_n^{{\tau }_k}\to X_n$ a.s. $k\to \mathrm{\infty }$ and:

$|X_n^{{\tau }_k}|$$\le {\displaystyle \sum _{j=0}^n}|X_j|\in {ℒ}^1\text{(}{ℒ}^1\text{-bound)}$

Thus $X_n^{{\tau }_k}\stackrel{\begin{array}{c}{L}^1\end{array}}{⟶}{X}_n,k\to \mathrm{\infty }\forall n\in \mathbb{N}$ by bounded convergence theorem.Hence $X$ must be martingale and we are done. ∎