I am trying to follow the proof of the following lemma from Rene Schilling's Brownian Motion and Stochastic Calculus. The setting is as below, where $\mathcal{L}_{T,loc}^2$ is the space of progressively measurable processes $f$ with a localizing sequence of stopping times $\tau_n$ that $\uparrow \infty$ a.s. and $f1_{[0,\tau_n)} \in \mathcal{L}_T^2$ for all $n \ge 0$. $\mathcal{L}_T^2$ is the progressively measurable $L^2$ functions in the measure space $\lambda_T \otimes P$, where $\lambda_T$ is the Lebesgue measure in $[0,T]$. In the following excerpt, simple processes refer to real-valued stochastic process $(f(t,\cdot))_{t \in [0,T]}$ of the form $$f(t,\omega)=\sum_{j=1}^n \phi_{j-1}(\omega)1_{[s_{j-1},s_j)}(t)$$where $n\ge 1$, $0=s_0\le s_1 \le \cdots \le s_n \le T$ and $\phi_j \in L^\infty (\mathcal{F}_{s_j})$ are bounded $\mathcal{F}_{s_j}$ measurable random variables, $j=0,\dots , n-1$. We write $\mathcal{S}_T$ for the family of all simple processes on $[0,T]$. We denote $\Pi$ as the partition of $[0,T]$ for the simple processes.
I have $2$ questions regarding the proof. First, how do we use a diagonal procedure to achive that the sequences $(b^\Pi)_\Pi, (\sigma^\Pi)_\Pi$ are independent of $n$? I cannot come up with an explicit construction of this.
Because usually to use diagonal argument, we would have, say the sequence of partitions $\Pi_2$ corresponding to $\tau_2$ be a subsequence of $\Pi_1$. But we don't have this here.
Finally, how can we calculate $$P\left( \sup_{t \le T} \left| \int_0^t (b_\Pi(s)-b(s)) \, ds \right|>\epsilon\right) \to_{|\Pi|\to 0} 0\text{?}$$
I tried imitating the proof for the case with $\sigma_\pi$, but the problem here is that we cannot use Doob's inequality as we don't have a martingale here. One thing I tried is to follow up to the third inequality as in the $\sigma$ case, and then bound \begin{align}
& P\left( \sup_{t \le T} \left| \int_0^{t\wedge \tau_n}(b_\Pi (s)-b(s)) \, ds \right| > \epsilon\right) \\[6pt]
\le {} & P\left(\sup_{t\le T} \int_0^{t \wedge \tau_n}| b_\Pi(s)-b(s)| \,ds>\epsilon\right) \\[6pt]
\le {} & P \left( \int_0^{T \wedge \tau_n}|b_\Pi (s)-b(s)| \, ds > \epsilon \right).
\end{align}
But I cannot really progress from here because if I use Chebyshev's inequality then I would need to take the square inside the integrand, which would require using Jensen's inequality but I'm not sure I can use this. This seems much more complicated than the previous calculation for $\sigma$. Is there any way to prove this? I would greatly appreciate some help.
Best Answer
For the diagonalization: Let's consider the diffusion coefficient $\sigma$ (the reasoning for the drift is analogous). Since $\sigma 1_{[0,\tau_n)} \in \mathcal{L}^2_T$ there exists for each a simple process $g_n$ such that
$$\|g_n- \sigma 1_{[0,\tau_n)} \|_{L^2} \leq \frac{1}{n}.\tag{1}$$
Proof: For each $n \geq k$ we have
\begin{align*}\|g_n 1_{[0,\tau_k)} - \sigma 1_{[0,\tau_k)}\|_{L^2}^2 &= \mathbb{E} \int_0^{\tau_k} |g_n(s,\omega)-\sigma(s,\omega)|^2 \, ds \, d\mathbb{P}(\omega) \\ &\leq \mathbb{E} \int_0^{\tau_n} |g_n(s,\omega)-\sigma(s,\omega)|^2 \, ds \, d\mathbb{P}(\omega) \\ &\leq \mathbb{E} \int_0^{\tau_n} |g_n(s,\omega)-\sigma(s,\omega)|^2 \, ds \, d\mathbb{P}(\omega) \\ &\quad +\mathbb{E} \int_{\tau_n}^{\infty} |g_n(s,\omega)-0|^2 \, ds \, d\mathbb{P}(\omega) \\ &= \|g_n- \sigma 1_{[0,\tau_n)}\|_{L^2}^2 \end{align*}
and so, by $(1)$,
$$\|g_n 1_{[0,\tau_k)} - \sigma 1_{[0,\tau_k)}\|_{L^2} \leq \frac{1}{n},$$ which proves the assertion. Consequently, $(g_n)_{n \in \mathbb{N}}$ is the sequence of simple functions which we are looking for.
Regarding your question about the estimate for the drift: Yes, you need to apply Jensen's inequality. Note that, by Jensen's inequality,
$$\left( \int_0^t f(s) \, ds \right)^2 \leq t \int_0^t f(s)^2 \, ds \tag{2}$$
for each $t \geq 0$ and any (suitable integrable) function $f$. This gives
\begin{align*} \left| \int_0^{T \wedge \tau_n} |b^{\Pi}(s)-b(s)| \, ds \right|^2 &\leq (T \wedge \tau_n) \int_0^{T \wedge \tau_n} |b^{\Pi}(s)-b(s)|^2 \, ds \\ &\leq T \int_0^{T \wedge \tau_n} |b^{\Pi}(s)-b(s)|^2 \, ds. \tag{3}\end{align*}
Taking expectation we get
\begin{align*} \mathbb{E}\left(\left| \int_0^{T \wedge \tau_n} |b^{\Pi}(s)-b(s)| \, ds\right|^2 \right)\leq T \mathbb{E}\int_0^{T \wedge \tau_n} |b^{\Pi}(s)-b(s)|^2 \, ds,\end{align*}
and by construction the right-hand side converges to $0$ as $|\Pi| \to 0$. Hence, by Markov's inequality,
\begin{align*} \mathbb{P} \left( \int_0^{T \wedge \tau_n} |b^{\Pi}(s)-b(s)| \, ds > \epsilon \right) &\leq \frac{1}{\epsilon^2}\mathbb{E}\left(\left| \int_0^{T \wedge \tau_n} |b^{\Pi}(s)-b(s)| \, ds \right|^2 \right) \\ &\xrightarrow[]{|\Pi| \to 0} 0. \end{align*}