Existence/uniqueness of the solution to some SDE with discontinuous coefficient Consider a SDE
$$dX_t = b(t,X_t)dt + f\big(a(t,X_t)\big)dW_t,\quad \quad\quad\quad\quad\quad\quad\quad\quad(\ast)$$
where $(W_t)_{t\ge 0}$ is a Brownian motion and
$$f(z):={\bf 1}_{\{z>0\}} +\frac{1}{2}{\bf 1}_{\{z\le 0\}}.$$
Assume $b, a: \mathbb R_+\times\mathbb R \to\mathbb R$ are both bounded and Lipschitz. Can we obtain the existence/uniqueness result for $(\ast)$?
Any answers, remarks or references are appreciated!
PS : I tried to approximate $(\ast)$ by replacing $f$ with
$$f_n(z):=\min\left(\frac{nx^++1}{2},~1\right),$$
where $x^+:=\max(x,0)$. Denote by $(n\ast)$ the corresponding SDE. Then it is not hard to show the existence/uniqueness of the solution, denoted by $X^n$, and further the tightness of $(X^n)_{n\ge 1}$. But I do not know how to the limit verifies $(\ast)$.
 A: I believe we can prove (strong) existence at least. I will take for granted a slightly modified version of the result proven by Mateusz Kwasniski here. Namely,
Lemma 1:
Let $W$ be a standard Brownian motion and $\sigma: [0, T] \times \mathbb R \to \mathbb R$ a Borel function with $C_1 < \sigma < C_2$ for some constants $C_1, C_2 > 0$.
Let $Y$ be the solution to the SDE
$$dY_t = \sigma(t, Y_t) dW_t$$ with $Y_0 = y_0$.
Fix $Y > 0$. Then there exists, for every $\varepsilon, h > 0$, a $\delta_{\varepsilon, h} > 0$ such that, for all $U \subset [0, T] \times \mathbb R$ with $\mu(U) < \delta$,
$$\mathbb P\left(\int_{0}^T \mathbf 1_{U} (s, X_s) ds > h\right) < \varepsilon.$$
Where $\mu$ denotes the (product) Lebesgue measure.
Proof of main theorem:
We work over the interval $[0, T]$, and without loss of generality assume for simplicity the initial condition $X_0 = 0$.
Let $X^n$ be the solution to the SDE
$$dX^n_t = b(t,X^n_t)dt + f_n \big(a(t,X^n_t)\big)dW_t$$
with $X^n_0 = 0$.
Where $f_n (z) := \frac{1}{2} + \frac{1}{2} \min(nx^+,1)$.
Define, for each $n$, a probability measure $\mathbb Q^n$ by $\frac{d\mathbb Q^n}{d\mathbb P} = \mathcal E(N_T)$, where $\mathcal E(.)$ denotes the Doleans Dade exponential, and $N_T := \int_{0}^T -\frac{b(t, X^n_t)}{f_n(a(t, X^n_t))} dB_t$.
Then by Girsanov's theorem, $\mathbb Q^n$ is equivalent to $\mathbb P$, and under $\mathbb Q^n$, we have that
$$dX^n_t = f_n \big(a(t,X^n_t)\big)dB^n_t$$
with $B^n_t$ a standard Brownian motion.
Let $\varepsilon, h > 0$ be arbitrary. We note also that $\frac{d\mathbb Q^n}{d\mathbb P}$ are uniformly integrable over $n$, thus there exists some $\eta > 0$ such that $\mathbb P(A) < \frac{\varepsilon}{2}$ whenever $\mathbb Q^n (A) < \eta$ for some $n$.
Choose some $K > 0$ large enough such that for all $n > 0$, $\mathbb P(\sup_t |X^n_t| \leq K) > 1 - \frac{\varepsilon}{2}$, which we may do by standard $L^p$ bounds on solutions to SDE with Lipschitz coefficients.
Now, for each $\delta > 0$, we may find some $N_{\delta} > 0$ such that $\mu((a^{-1}((0, \frac{1}{N_{\delta}}))) \cap ([0, T] \times [-K, K])) < \delta$.
Thus, for all $n$, we have $\mathbb Q^n \left(\{\int_{0}^T \mathbf 1_{(0, 1/N)} (a(s, X^n_s))ds > h \} \text{ and } \{\sup_t |X^n_t| < K\}\right)< \eta$ as soon as we set $\delta = \delta_{\eta, h}$ as in Lemma 1, and $N = N_{\delta}$. Consequently, $\mathbb P\left(\int_{0}^T \mathbf 1_{(0, 1/N)} (a(s, X^n_s))ds > h \right)< \varepsilon$. $(*)$
Admitting the ucp convergence of $X^n$ to $X$ as stated in the original post, we now show that the limit $X$ satisfies the SDE given. We compute
$|X_t - \int b(t,X_t) dt - \int f\big(a(t,X_t)\big)dW_t|$
$\leq |X_t - X^n_t| + |X^n_t - \int b(t,X_t) dt - \int f\big(a(t,X_t)\big)dW_t|$
$= |X_t - X^n_t| + |\int b(t,X^n_t) dt + \int f_n\big(a(t,X^n_t)\big)dW_t - \int b(t,X_t) dt - \int f\big(a(t,X_t)\big)dW_t|$
$\leq |X_t - X^n_t| + |\int b(t,X^n_t) dt - \int b(t,X_t) dt| + |\int f_n\big(a(t,X^n_t)\big)dW_t - \int f \big(a(t,X^n_t)\big)dW_t| + |\int f \big(a(t,X^n_t)\big)dW_t - \int f\big(a(t,X_t)\big)dW_t|.$
Since $X$ is the ucp limit of the $X^n$, the first term goes to $0$ in probability uniformly in $t$, while the same goes for the second and fourth terms by the dominated convergence theorem for stochastic integrals.
For the third term, we crucially use $(*)$ to obtain ucp convergence of $\int f_n\big(a(t,X^n_t)\big) dW_t$ to $\int f \big(a(t,X^n_t)\big) dW_t$, noting that $f = f_n$ outside of $(0, \frac{1}{n})$.
Let us show in more detail the claimed convergence for the third and fourth term.
Third term:
Let $\varepsilon, \delta > 0$ be arbitrary. We will show that $\mathbb P \big(\mathbf 1_{A_n^c} \int_{[0, T]} f_n\big(a(t,X^n_t)\big) - f \big(a(t,X^n_t)\big)dW_t > \delta\big)< \varepsilon$ for all large enough $n$.
To this end, fix some $h > 0$ depending on $\varepsilon, \delta$ to be chosen later.
By $(*)$, there exists $N > 0$ such that $\mathbb P\left(\int_{0}^T \mathbf 1_{(0, 1/N)} (a(s, X^n_s))ds > h\right)< \frac{\varepsilon \sqrt \delta}{2T}$ for all $n \in \mathbb Z_+$.
Consequently, denoting by $A_n$ the event $\{\int_{0}^T \mathbf 1_{(0, 1/N)} (a(s, X^n_s))ds > h \}$, for all $n > N$ we have
$\mathbb E \big[\left(\int_{[0, T]} f_n\big(a(t,X^n_t)\big) - f \big(a(t,X^n_t)\big)dW_t\right)^2\big]$
$= \mathbb E \left[\int_{[0, T]} \left[f_n\big(a(t,X^n_t)\big) - f\big(a(t,X^n_t)\big)\right]^2 dt\right]$
$= \mathbb E \left [\mathbf 1_{A_n^c}\int_{[0, T]} \left[f_n\big(a(t,X^n_t)\big) - f\big(a(t,X^n_t)\big)\right]^2 dt\right] + \mathbb E \left[[\mathbf 1_{A_n}\int_{[0, T]} \left[f_n\big(a(t,X^n_t)\big) - f\big(a(t,X^n_t)\big)\right]^2 dt\right]$
$=  \mathbb E\left[\mathbf 1_{A_n^c} \int_{t \in S} \left[f_n\big(a(t,X^n_t)\big) - f\big(a(t,X^n_t)\big)\right]^2 dt\right] + \mathbb E\left[\mathbf 1_{A_n^c} \int_{t \in S^c} \left[ f_n\big(a(t,X^n_t)\big) - f\big(a(t,X^n_t)\big)\right]^2 dt\right] + \mathbb E \left[[\mathbf 1_{A_n}\int_{[0, T]} \left[f_n\big(a(t,X^n_t)\big) - f\big(a(t,X^n_t)\big)\right]^2 dt\right]$
where here $S$ denotes the (random) set of all $t \in [0, T]$ such that $a(s, X^n_t) \in (0, \frac{1}{N})$. Noting now that noting that $f = f_n$ outside of $(0, \frac{1}{N})$, we have that the integral over $S^c$ is zero. Consequently, continuing the manipulation from above we have
$\dots = \mathbb E\left[\ \mathbf 1_{A_n^c} \int_{t \in S} \left[f_n\big(a(t,X^n_t)\big) - f\big(a(t,X^n_t)\big)\right]^2 dt\right] + \mathbb E \left[[\mathbf 1_{A_n}\int_{[0, T]} \left[f_n\big(a(t,X^n_t)\big) - f\big(a(t,X^n_t)\big)\right]^2 dt\right]$
$\leq \mathbb E \left[\mathbf 1_{A_n^c} \int_{t \in S} \frac{1}{2}dt\right] + E[\frac{T}{2}\mathbf 1_{A_n}]$
= $\frac{h}{2} + \frac{\varepsilon \delta^2}{2}.$
And thus by the Markov inequality we have
$\mathbb P \big(\left [ \int_{[0, T]} f_n\big(a(t,X^n_t)\big) - f \big(a(t,X^n_t)\big)dW_t\right]^2 > \delta^2\big) < \frac{\varepsilon \delta^2 + h}{2\delta^2}$.
Setting now $h = \varepsilon \delta^2$, we obtain
$\mathbb P \big(\left[\mathbf \int_{[0, T]} f_n\big(a(t,X^n_t)\big) - f \big(a(t,X^n_t)\big)dW_t\right]^2 > \delta^2 \big) < \varepsilon$
and thus
$\mathbb P \big(\left|\mathbf \int_{[0, T]} f_n\big(a(t,X^n_t)\big) - f \big(a(t,X^n_t)\big)dW_t\right| >  \delta\big)< \varepsilon$
as claimed.
Fourth term:
Let again $\varepsilon, \delta > 0$ be arbitrary. Fix $h, r > 0$ depending on $\varepsilon, \delta$ to be chosen later.
By the same reasoning as $(*)$, there exists $N > 0$ such that $\mathbb P\left(\int_{0}^T \mathbf 1_{(-1/N, 1/N)} (a(s, X^n_s))ds > h\right) < r$ for all $n \in \mathbb Z_+$. For each $n$, denote by $A_n$ the event $\{\int_{0}^T \mathbf 1_{(-1/N, 1/N)} (a(s, X^n_s))ds > h\}$.
By the ucp convergence of $X^n$ to $X$, and the uniform continuity of $a$, we may choose $N_0$ large enough such that for each $n > N_0$, $|a(s, X^n_s) - a(s, X_s)| < \frac{1}{2N}$ on an event $B_n$ with $\mathbb P(B_n) >1 - r$.
Thus on $A^c_n \cap B_n$ we have that for all $s$ in a (random) set $H(n, \omega)$ of measure greater than $T - h$, either both $a(s, X^n_s)$ and $a(s, X_s)$ are less than $\frac{1}{2N}$, or both are greater than $\frac{1}{2N}$. We conclude that $\mathbb P \big(\big \{\mu\left(\left \{s \in [0, T]| \ f\left(a(s, X^n_s)\right) \neq f\left(a(s, X^n_s)\right) \right \}\right) > h\big \} \big) < 2r$.
Thus we estimate for all $n > \max(N, N_0)$,
$\mathbb E \big [\left (\int_{[0, T]} f \big(a(t,X^n_t)\big) - f \big(a(t,X_t)\big)dW_t\right)^2\big ]$
$=\mathbb E \big [\int_{[0, T]}\big( f\left(a(t,X^n_t)\right) - f \left(a(t,X_t)\right)\big)^2dt \big ]$
$=\mathbb E \big [ \mathbf 1_{A^c_n \cap B_n}  \int_{[0, T]}\big( f\left(a(t,X^n_t)\right) - f \left(a(t,X_t)\right)\big)^2dt \big ]+  \mathbb E \big [ \mathbf 1_{(A^c_n \cap B_n)^c} \int_{[0, T]}\big( f\left(a(t,X^n_t)\right) - f \left(a(t,X_t)\right)\big)^2dt \big]$
The first term is $0$ on $H(n, \omega)$ as stated earlier, so we have
$… = \mathbb E \big [ \mathbf 1_{A^c_n \cap B_n}  \int_{[0, T] \setminus H(n, \omega)}\big( f\left(a(t,X^n_t)\right) - f \left(a(t,X_t)\right)\big)^2dt \big ] + \mathbb E \big [ \mathbf 1_{(A^c_n \cap B_n)^c} \int_{[0, T]}\big( f\left(a(t,X^n_t)\right) - f \left(a(t,X_t)\right)\big)^2dt \big]$
$ \leq \mathbb E \big [ \mathbf 1_{A^c_n \cap B_n}  \int_{[0, T] \setminus H(n, \omega)}\frac{1}{4} dt \big ] + \mathbb E \big[ \mathbf 1_{(A^c_n \cap B_n)^c} \int_{[0, T]} \frac{1}{4} dt \big]$
$= \frac{h}{4} + \frac{rT}{2}.$
Thus by the Markov inequality we have
$\mathbb P \big (\left (\int_{[0, T]} f \big(a(t,X^n_t)\big) - f \big(a(t,X_t)\big)dW_t\right)^2 > \delta^2 \big) < (\frac{1}{\delta^2}) (\frac{h}{4} + \frac{rT}{2})$.
Setting now $h = 2\varepsilon \delta^2$ and $ r = \frac{\varepsilon \delta^2}{T}$, we have
$\mathbb P \big (\left (\int_{[0, T]} f \big(a(t,X^n_t)\big) - f \big(a(t,X_t)\big)dW_t\right)^2 > \delta^2 \big) < \varepsilon$
so that $\mathbb P \big (\left (\int_{[0, T]} f \big(a(t,X^n_t)\big) - f \big(a(t,X_t)\big)dW_t\right)^2 > \delta \big) < \varepsilon$
as required.
Conclusion:
Thus $X$ satisfies the SDE, and so we have strong existence as claimed.
