I am currently going through the proof of the existence of a solution of the SDE
\begin{align} dX_t = bdt + \sigma dB_t \end{align}
where $B_t$ is a Brownian motion with respect to a filtration $\{\mathcal{F}_t\}_{t\in[0,T]}$. We assume the following conditions hold for $t \in [0,T]$:
1) $b(t,X_t) \text{ and } \sigma(t,X_t) \text{ are } \mathcal{B}\times\mathcal{F} \text{-measurable}$ and $\mathcal{F}_t$-adapted.
2) $|b(t,x)| \leq L(1 + |x|)$ and $\sigma(t,x) \leq L(1 + |x|)$.
3) $|b(t,x) - b(t,y)| \leq L|x - y|$ and $|\sigma(t,x) - \sigma(t,y)| \leq L|x - y|$.
4) $X_0$ is $\mathcal{F}_0$-measurable, and $E[|X_0|^2] < \infty$.
Now, every proof I have seen uses the Picard iterations
\begin{align} \begin{cases} X_t^0 &= X_0. \\ X_t^{n+1} &= X_0 + \int_0^tb(s,X_s^n)ds + \int_0^t \sigma(s,X_s^n)dB_s. \end{cases} \end{align}
There are several things I am struggling to understand in the proof, and I am happy with any help/hints I can get.
First we need to show that the iterates $X_t^n$ are elements of $L_{\text{ad}}^2([0,T]\times \Omega)$, that is, that properties a), b), and c) are satisfied:
a) $X_t^n$ is continuous in $t$
b) $X_t^n$ is $\mathcal{B}\times\mathcal{F}$ -measurable and $\mathcal{F}_t$-adapted
c) $E\Big[ \int_0^t |X_s^n|^2 ds \Big] < \infty$
We proceed by induction, and clearly a) - c) are satisfied for $n=0$, due to condition 4). Now, we assume that a) - c) hold for $X_t^n$. By condition 2), the Cauchy-Schwarz(C-S) inequality as well as $(x_1+x_2+ \cdots+x_n)^2 \leq n(x_1^2 + x_2^2 + \cdots + x_n^2)$ it is easy to see that
\begin{align*} E\bigg[ \int_0^t \big| \sigma(s,X_s^{n})\big|^2 ds \bigg] &\leq 2 L^2 \int_0^t 1 + E\big[ |X_s^{n}|^2 \big] ds < \infty \\ \int_0^t \big| b(s,X_s^{n}) \big| ds & \leq L\sqrt{2t} \sqrt{\int_0^t 1 + |X_s^{n-1}|^2 ds} < \infty \end{align*} where the last inequality holds almost surely. Hence the above integrals exist with probability 1. Now, the argument is that this implies that properties a) and b) are satisfied. I fail to see why.
Thoughts:
$T_1$ - If $\sigma(t,X_t^n)$ is measurable and adapted, we have that, since the expected integral of $\sigma(t,X_t^n)^2$ is finite, that the Ito integral is $t$-continuous a.s. (a property of processes satisfying a) - c) ).
$T_2$ - Once a) and b) is settled, c) is found by using condition 2), $(x_1+x_2+ \cdots+x_n)^2 \leq n(x_1^2 + x_2^2 + \cdots + x_n^2)$ and C-S.
Questions:
$Q_1$ - We know from condition $1$) that for instance $b(t,X_t)$ is jointly measurable and adapted. Is this different from $b(t,X_t^n)$ satisfying these properties?
I really don't see how to prove that $X_t^{n+1}$ is jointly measurable and adapted.