On Tanaka's SDE Throughout this question, for an arbitrary real valued continuous time stochastic process $Y$, I define $\{\mathcal{F}_t^Y\}_{t \ge 0}$ to be the filtration generated by $Y$.  I also define $\text{sgn}(x) = \mathbf{1}(x > 0) - \mathbf{1}(x \leq 0)$
The standard argument that Tanaka's SDE has no strong solutions for a particular probability space is as follows, but I do not understand the last part of it: it suffices to consider a probability space $(\Omega, \mathcal{F}, P)$ on which there exists a Brownian motion $B$, and we define our filtration $\{\mathcal{F}_t\}_{t \ge 0}$ to be that which is generated by $B$.  If $$dX_t = \text{sgn}(X_t) dB_t $$ has a strong solution, then $$B_t = \int_0^t \text{sgn}(X_s)dX_s = |X_t| - L_t^0$$ where $L_t^0$ is the local time of $X$ about $0$ at time $t$.  It is well known that $$L_t^0 = |X_t| - \lim_{h \rightarrow 0} \frac{1}{h}\text{Leb}\left( \{s \in [0,t] : \left|X_s \right| \leq h  \right)$$ and thus $$\mathcal{F}_t^{X} \subseteq \mathcal{F}_t^B \subseteq \mathcal{F}_t^{|X|}$$ which is a contradiction, since $$\text{sgn}(X_t) \text{ cannot possibly be } \mathcal{F}_t^{|X|} \text{ measurable.} \qquad \qquad \textbf{(1)}$$ Intuitively, $\textbf{(1)}$ is true, but how does one rigorously go about showing this?
I try to do so by contradiction, but it leads me nowhere after a bit.  If $\text{sgn}(X_t)$ were $\mathcal{F}_t^{|X|}$ measurable, then there would exist some functional $f$ defined on the space of continuous functions on $[0,t]$ such that $$\text{sgn}(X_t) = f(|X_s|, s \leq t) $$ I know this is intuitively obvious to be a contradiction, but I cannot formally show it.  Any help would be appreciated!
 A: So my original answer did not take into account the fact that there could exist a transformation of the entire path $(|X_{s}|)_{0\leq s\leq t}$. Instead, I think I have found a way to prove that for a Brownian motion $X$, we cannot have the inclusion $\mathcal{F}_{t}^{X}\subseteq\mathcal{F}_{t}^{|X|}$. Since the solution to Tanaka's SDE is a Brownian motion, this will suffice.
Define the $\sigma$-algebra
$$
\mathcal{D}_{t} = \lbrace F\in \mathcal{F}_{t}^{|X|}\mid \mathbb{E}\left[ X_{t}\mathbb{1}_{F} \right] = 0 \rbrace
$$
That this is in fact a $\sigma$-algebra follows by linearity of the integral and that $X_{t}\in L^{1}$ with expectation 0.
Let $B\in\mathcal{B}(\mathbb{R})$ and $s\leq t$. Then $\lbrace |X_{s}|\in B\rbrace \in \mathcal{D}_{t}$ since by the reflection principle
$$
\mathbb{E}\left[ X_{t}\mathbb{1}_{\lbrace |X_{s}|\in B\rbrace } \right] = 0
$$
This means that $\mathcal{G}_{t}=\mathcal{F}_{t}^{|X|}$, at least if we assume the filtration $\mathcal{F}_{t}^{|X|}$ to be complete.
But since
$$
\mathbb{E}\left[ X_{t}\mathbb{1}_{\lbrace X_{t}>0\rbrace} \right] > 0
$$
we have $\lbrace X_{t}>0\rbrace\notin\mathcal{F}_{t}^{|X|}$, but naturally we have $\lbrace X_{t}>0 \rbrace\in\mathcal{F}_{t}^{X}$ and thus we must have
$$
\mathcal{F}_{t}^{X}\not\subseteq\mathcal{F}_{t}^{|X|}
$$
as desired.
This lines up with the intuition that knowing something about $|X_{t}|$ does not tell us about the sign of $X_{t}$.
