I have some questions about the proof of Doob's upcrossing estimation.
Lemma
Be $(X_n)$ a submartingale and be $U([a, b]; N; x)$ the number of upcrossings of $(X_n(x))$ across $[a, b]$ running at $0 \leq n \leq N$, where $n, N \in \mathbb{N}$. Then $$\int_A U([a, b];N;x)d\mathbb{P} \leq \frac{1}{b-a}\int_A (X_N - a)^+ d\mathbb{P}$$ $\forall A \in \mathcal{F}_0$.
- First of all: why we need $A\in\mathcal{F}_0$? We are working in $(\Omega, \mathcal{F}, \mathcal{F}_n, \mathbb{P})$ where $\mathcal{F}_n$ is a filtration, that is an increasing sequence of sub $\sigma$ algebras such that $\mathcal{F}_0 \subset \mathcal{F}_1 \ldots \subset \mathcal{F}$.
What I know is that the space $(\Omega, \mathcal{F}, \mathcal{F}_n, \mathbb{P})$ is a probability space called "filtered and sigma-finite" that is: an exhaustive sequence $A_n \uparrow \Omega$ exists and $\mathbb{P}(A_n) \leq +\infty)$. It's not clear to me why $A \in \mathcal{F}_0$...
Proof
We introduce the following stopping time
$$S_k := \min\{ \inf \{ n > T_{k-1}; X_n \leq a \}, N \}$$ $$T_k := \min\{ \inf \{ n > S_k; X_n \geq b \}, N \}$$
(I proved they are indeed stopping times). Shortcut: $U(\ldots) = U$.
Now we can see that:
$$T_0 = 0 < S_1 < T_1 < S_2 < \ldots < S_U < T_U \leq S_{U+1} \leq T_{U+1} = \ldots = S_N = T_N = N$$
- Here it's not clear to me what $S_U, T_U$ really mean. Also I don't get why from $U+1$ onwards, $S$ and $T$ are identical to $N$ (up to $S_N = T_N = N$)
Now "by definition of upcrossing" (unclear):
$$(b-a)U \leq (X_{T_1} - a) + (X_{T_2} - X_{S_2}) + \ldots + (X_{T_U} - X_{S_U})$$
where all the brackets are $\geq (b-a)$
- (why exactly?)
Then
$$-(X_n - a)^- \leq (X_{T_{U+1}} - X_{S_{U+1}}) + (X_{T_{U+2}} - X_{S_{U+2}}) + \ldots (X_{T_{N}} - X_{S_{N}})$$
Where the brackets with indexes greater than $U+1$ are all zero (this makes sense with respect to what we defined upon).
Good, now we subtract the second from the first and we integrate, and here I have no problems since I understood that
$$ \begin{align} (b-a) \int_A U d\mathbb{P} & - \int_A (X_n - a)^- d\mathbb{P} \leq \\\\ & \leq - \int_A a d\mathbb{P} + \int_A (X_{T_1} - X_{S_2})d\mathbb{P} + \ldots + \int_A (X_{T_{N-1}} - X_{S_N}) d\mathbb{P} + \int_A X_{T_N}d\mathbb{P} \\\\ & \leq \int_A (X_{T_N} - a) d\mathbb{P} \end{align} $$
Hence
$$\int_A U d\mathbb{P} \leq \frac{1}{b-a}\int_A \big((X_N - a)^- + (X_{T_N} - a)\big) d\mathbb{P}$$
Now $X_{T_N} = X_N$ hence using generic $Q = Q^+ + Q^-$ we obtain the result.
So can you please clarify my doubts? I know there are many but I really need to understand those things, I cannot learn by heart.