A note on Doob's theorem

Question

I have faced the following problem, regarding to the Martingale Theory. Because this area far from my area I don't know whether this problem is in literature or this can be simple question for specialists of this area. If like this things in any literature can someone suggest me?

Let $$ (\mathbb{P}_{n}(t))_{n \in \mathbb{N}_{0}} = \left( \left\{ I^{n}_{j}(t),~ 1\leq j \leq q_{n}(t) \} \right\} \right)_{n \in \mathbb{N}_{0}} $$ be a sequence of partitions of $ [0,1)$ depending on parameter $t\in J\subset (0,1)$ such that $$ \underset{t\in J}{\sup}\underset{1\leq j\leq q_{n}(t)}{\max}|I^{n}_{j}(t)|\leq \lambda ^{n} $$ for some $\lambda \in (0,1)$ and for any $t\in J$ we have $q_{n}(t)\rightarrow \infty.$

Suppose that $ f $ is $ 1 $-periodic and that $ f \in {L^{p}}([0,1) $, where $ p > 1 $. Next, for any $t\in J,$ we define a sequence $ (G_{n}(\cdot, t): [0,1) \to \mathbb{C})_{n \in \mathbb{N}_{0}} $ of functions by $$ \forall n \in \mathbb{N}_{0}, ~ \forall x \in [0,1), ~ \forall j \in \{ 1,\ldots, q_{n}(t) \}: \\ {G_{n}}(x, t) \stackrel{\text{df}}{=} \frac{1}{\left| I^{n}_{j}(t) \right|} \int_{I^{n}_{j}(t)} f(s) ~ \mathrm{d}{s}, \quad \text{if $ x \in I^{n}_{j}(t) \in D_{n}(t) $}. $$ Assume that for any $t\in J$ the random value $G_{n}(\cdot, t)$ is a $L^{p}$ bounded martingale w.r.t $\mathbb{P}_{n}(t).$ It is well known (according to Doob's theorem) $$ G_{n}(\cdot, t)\rightarrow f, \,\,\,a.s. $$ and $$ E(|G_{n}(\cdot, t)-f|^{p})\rightarrow 0. $$

Question. Is it true $$ \underset{t\in J}{\sup}E(|G_{n}(\cdot, t)-f|^{p})\rightarrow 0? $$

Since the lengths of atoms of partitions uniformly tend to zero, I think it is true, but I do not know how to show it.

Answer 1

It is true. Here is an elementary proof, without relying on the martingale convergence theorem.

Let us first consider the case when $f:[0,1]\to \mathbb{R}$ is continuous. If $I\subseteq [0,1]$ is an interval of length $|I|$ which contains $x$, define the modulus of continuity of $f$ as $$m_f(\delta):=\sup\{| f(s) - f(t)|: s,t\in [0,1]: |s-t|<\delta\}, \quad \delta>0,$$ and notice that $m_f:(0,1)\to [0,\infty)$ is increasing and $\lim_{\delta \downarrow 0} m_f(\delta) = 0$, since $f$ is uniformly continuous on $[0,1]$. The bound $$\Big|f(x)-\frac{1}{|I|}\int_{I} f(s)ds \Big|\leq m_f(\delta),$$ clearly holds, and implies that $$\| f - G_n(t,\cdot)\|_{L^p} \leq \sup_{x\in [0,1]}|f(x) - G_n(t,x)|\leq m_{f}(\lambda^n) \quad \forall t\in J, $$ and so in this case the thesis follows from $\lim_{\delta \downarrow 0} m_f(\delta) = 0$ and $0<\lambda^n \to 0$.

For later use notice that, since $G_n(t,\cdot)$ is the conditional expectation $E[f|\mathcal{F}_n(t)]$ of $f$ with respect to the sigma algebra $\mathcal{F}_n(t)$ generated by the partition $\mathbb{P}_n(t)$ (using the Lebesgue measure on $[0,1]$ as the underlying probability), we have proved that $$\sup_t \|f- E[f|\mathcal{F}_n(t)]\|_{L^p}\to 0 \text{ as } n\to \infty, \quad \text{ for any continuous $f$}.$$

To treat the case of a general $f\in L^p$, we approximate it with a continuous $g$. So, choose $\epsilon>0$ and find $g:[0,1]\to \mathbb{R}$ continuous such that $ \| f - g\|_{L^p} \leq \epsilon $. Since the conditional expectation is a linear contraction in $L^p$ we have that $$\|E[f|\mathcal{F}_n(t)]- E[g|\mathcal{F}_n(t)]\|_{L^p}\leq \| f -g \|_{L^p} \quad \text{ for all } t\in J.$$ Thus, adding and subtracting $g- E[g|\mathcal{F}_n(t)]$ from $f- E[f|\mathcal{F}_n(t)]$ and applying the triangle inequality we find that $$\|f- E[f|\mathcal{F}_n(t)]\|_{L^p}\leq 2\| f -g \|_{L^p}+\|g- E[g|\mathcal{F}_n(t)]\|_{L^p}.$$ Since $g$ is continuous, as we proved above there exist $k$ s.t. $$\sup_t \|g- E[g|\mathcal{F}_n(t)]\|_{L^p}\leq \epsilon\quad \text{ for all } n\geq k, $$ and thus $\sup_{t\in J}\|f- E[f|\mathcal{F}_n(t)]\|_{L^p}\leq 3\epsilon $ for all $n\geq k$, proving the thesis.