Let $U_n\sim$ Unif$(-1,1)$ and $\Sigma_n=\sigma(U_1,...,U_n)$. Define $X_0=0$ and $$X_n=X_{n-1}+(1-|X_{n-1}|)U_n.$$
Given that for the limit $X$ of $X_n$ we have $\mathbb{P}(X=-1)=\mathbb{P}(X=1)=\frac{1}{2}$, how do we show that $\{X=1\}\notin\Sigma_n$?
To show that we have a.s. and $L_1$ convergence of $X_n$ to $X$ we can show uniform integrability. To do this I thought we can show that $|X_n|\leq 1$ for all $n$ a.s. How do I do this?
If we have $L_1$ convergence then
$$\mathbb{E}[X \mid \Sigma_n]=X_n.$$
Further we have $\int_{\{X=1\}}X \, d\mathbb{P}=\mathbb{P}(X=1)=\frac{1}{2}$.
If $\{X=1\}\in\Sigma_n$, then we would have
$$\int_{\{X=1\}}X \, d\mathbb{P}=\int_{\{X=1\}}\mathbb{E}[X \mid \Sigma_n] \, d\mathbb{P}=\int_{\{X=1\}}X_n \, d\mathbb{P}.$$
Am I approaching this in the right way? How do I continue?
Suppose that $\{X=1\} \in \Sigma_n$ for some $n \in \mathbb{N}$. Since $\Sigma_n$ is a $\sigma$-algebra this would imply $$\{X=-1\} = \{X=1\}^c \in \Sigma_n.$$ Using
$$X = 1_{\{X=1\}} - 1_{\{X=-1\}}$$
it follows that
$$\mathbb{E}(X \mid \Sigma_n) = X.$$
On the other hand, the martingale property also gives
$$\mathbb{E}(X \mid \Sigma_n) =X_n,$$
and so $X_n = X$ almost surely. In particular, $$\mathbb{P}(X_n=1) = \mathbb{P}(X_n = -1) = \frac{1}{2}. \tag{1}$$ Since the mapping $x \mapsto f(x) := x+(1-|x|)u$ satisfies
$$f(x)=1 \iff x=1 \quad \text{and} \quad f(x)=-1 \iff x=-1$$
for any fixed $u \in [-1,1]$, we have
$$X_n(\omega) = 1 \iff X_{n-1}(\omega)=1 \quad \text{and} \quad X_{n}(\omega)=1 \iff X_{n-1}(\omega)=-1$$ Combining this with $(1)$ we get $$\mathbb{P}(X_{n-1}=1) =\mathbb{P}(X_{n-1} = -1) = \frac{1}{2}.$$ Iterating the procedure we find that $$\mathbb{P}(X_0=1) = \mathbb{P}(X_0=-1) = \frac{1}{2}$$ in contradiction to the assumption that $X_0 =0$.