Here is the doob's convergence theorem:
If $(X_{n})$ is a sup-MG such that $\sup_{n}\mathbb{E}(X_{n})_{-}<\infty$, then $X_{n}\longrightarrow X_{\infty}$ a.s.
There is also a version for $(X_{n})$ sub-MG, only alter $\sup_{n}\mathbb{E}(X_{n})_{+}<\infty$.
My confusion is about the requirement: $\sup_{n}\mathbb{E}(X_{n})_{-}<\infty$ for sup-MG and $\sup_{n}\mathbb{E}(X_{n})_{+}<\infty$ for sub-MG.
What does this requirement mean? Some note direct use $\sup_{n}\mathbb{E}|X_{n}|<\infty$, but is this redundant? Since no matter what you have, a MG, sub-MG or a sup-MG. Being one of them, $(X_{n})$ must be integrable, $\sup_{n}\mathbb{E}|X_{n}|<\infty$ is immediate.
So it must be the case that I interpret this requirement incorrectly — this requirement is not about the integrability.
But what does it imply if it is not about integrability?
Thank you!
Edit 1:
Okay, I've finished final and had some days off.
I just changed the title of this post, and as the new title said, we need to talk about the $L_{p}$ bounded, instead of integrability. As George pointed out, the condition $\sup_{n}|X_{n}|<\infty$ is stronger than integrability.
I am gonna call it $L_{1}$ bounded.
Note that the condition $\sup_{n}|X_{n}|<\infty$ is for Martingale. For sub-Martingale, we require to have $\sup_{n}(X_{n})_{+}<\infty$, and for super-Martingale, we require it to have $\sup_{n}(X_{n})_{-}<\infty$.
Similarly, for Doob's $L_{p}$ convergence theorem, we need the Martingale to have $\sup_{n}|X_{n}|^{p}<\infty$, $p\geq 2$. However, Doob's $L_{p}$ convergence only holds for Martingale, so there is no alternative condition for sub-Martingale or super-Martingale. I will call this condition as $L_{p}$ bounded.
George has given some examples showing this $L_{p}$ bounded requirement is stronger than integrability, and I will give an example showing why this requirement has been added into the convergence theorem. Namely,
Provide an example of a martingale which converges a.s to $-\infty$, but this example does not contradict Doob's Convergence Theorem.
Consider $S_{n}:=\sum_{k=1}^{n}X_{k}$ with $(X_{k})$ independent and satisfying $$\mathbb{P}(X_{k}=k^{2}-1)=\dfrac{1}{k^{2}}\ \text{and}\ \mathbb{P}(X_{k}=-1)=1-\dfrac{1}{k^{2}}.$$
Firstly, $S_{n}$ clearly adapts to the canonical filtration $\mathcal{F}_{n}^{X}:=\sigma(X_{0},\cdots, X_{n})$, and since $$\mathbb{E}|X_{k}|=2\Big(1-\dfrac{1}{k^{2}}\Big)<\infty,$$ it is clear to see that $$\mathbb{E}|S_{n}|<\infty.$$ Then, note that $\mathbb{E}X_{k}=0$, and using independence and taking-out-known property, we have $$\mathbb{E}(S_{n+1}|\mathcal{F}_{n}^{X})=S_{n}+\mathbb{E}(X_{n+1}|\mathcal{F}_{n}^{X})=S_{n}+\mathbb{E}(X_{n+1})=S_{n}.$$
However, $$\sum_{k=1}^{\infty}\mathbb{P}(X_{k}=k^{2}-1)=\sum_{k=1}^{\infty}\dfrac{1}{k^{2}}<\infty,$$ so Borel Cantelli I implies $$\mathbb{P}(X_{k}=k^{2}-1\ \text{i.o.})=0,$$ which in turns implies that $$\mathbb{P}(X_{k}=-1\ \text{e.v.})=1.$$
This means, the sequence $(X_{k})$ only finitely many term that is NOT $-1$, and starting at some $N$ large enough, $X_{n}=-1$ for all $n>N$. Therefore, your $S_{n}$, if $n$ is sufficient large, will equivalently be an infinite sum of $-1$, and thus $S_{n}$ converges almost surely to $-\infty$.
However, note that we have showed $S_{n}$ is a martingale, meaning it must be integrable, so what is happening here? Why does NOT $S_{n}$ converge almost surely to a finite limit, which is what Doob's Convergence Theorem says?
This is where we need the Martingale to be $L_{1}$ bounded. In fact, the $S_{n}$ in our example is NOT $L_{1}$ bounded.
Indeed, we will show $\mathbb{E}(S_{n})_{-}$ is NOT bounded. Take $\ell$ large enough but finite such that $$\sum_{k=\ell+1}^{\infty}k^{-2}\leq \dfrac{1}{2},$$ and set $$b:=\sum_{k=1}^{\ell}k^{2}.$$
Then with probability at LEAST $1/2$, $X_{k}=-1$ for all $k>\ell$ and hence with probability at least $1/2$, $S_{n}\leq b-n$.
This implies that $$\mathbb{E}(S_{n})_{-}\geq\dfrac{1}{2}(n-b)_{+}\longrightarrow\infty,\ \text{as}\ n\longrightarrow\infty.$$
This example shows that the $L_{p}$ bounded is essential to allow you have a finite limit in the almost sure convergence.
It seems that the $L_{p}$ bounded requirement indeed increases the difficulty when you want to apply Doob's Convergence.
In my personal experience, you only need to show the limit of $\mathbb{E}|X_{n}|$ or $\mathbb{E}|X_{n}|^{p}$ exists. To show this, you may need to write the expectation explicitly and then use some bounded and regular convergence theorem. (Like monotone convergence theorem and dominated convergence theorem)
Hope you guys enjoy George's and mine example 🙂 Hope my professor can give me a good grade since the final was really hard TAT.
Best Answer
Note that $(X_n)_{-} = \max(0,-X_n)$, the negative component of $X_n$. $(X_n)_{+} = \max(0,X_n)$.
Having integrability of these processes is less than asking that $|X_n|$ is integrable.
requiring $\sup E[(X_n)_{\pm}]<\infty$ can actually be stronger, because $E[|X_n|]$ can be integrable for each $n$, but converge to $\infty$ as $n \to \infty$. Take for example if $E|X_n| = |n|$ but $ \sup E[(X_n)_{+}] = \infty$.
Another example: $E|X_n| = \infty$ with all of the mass coming from $E[(X_n)_{+}] = \infty$, but $E[(X_n)_{-}] < \infty$ and $X_n$ a super martingale. Then we can still apply the theorem.