I have an i.i.d sequence $(u_j)_{j\in \mathbb{Z}_+}$ with zero mean and finite variance, say $\sigma^2$. Furthermore, I have another random variable $X_0$ (defined on the same probability space) which is independent of $(u_j)_{j\in \mathbb{Z}_+}$. It is also given that $E[X_0^2] < \infty$. Finally, I have a real-valued constant $\theta$ with $\lvert\theta\rvert < 1$.
Here is the recursive relation mentioned in the title. $$X_j = \theta X_{j-1} + u_j \quad j=1,2,\ldots$$
I want to show that $(X_{j}^2)_{j\in \mathbb{Z}_+}$ is uniformly integrable. Since higher moments ($>2$) of $u_1$ do not necessarily exist, the road to uniform integrability via $L^p$-boundedness ($p>1$) is blocked. Can someone help out?
Define $c_j:=\mathbb E\left[X_j^2\right]$. By the assumptions, we have $$c_j=\theta^2c_{j-1}+\sigma^2,\quad j\geqslant 1,$$ hence taking $R\geqslant\max\left\{\sigma^2/(1-\theta^2),\mathbb E\left[X_0^2\right]\right\}$, we can see by induction that $c_j\leqslant R$ for each $j\geqslant 0$.
Now, using boundedness of $\left(c_j\right)_{j\geqslant 1}$ and the recursion relation, that for each $A$, we have $$\sup_{j\geqslant 1}\mathbb E\left[X_j^2\mathbf 1(A)\right]\leqslant \frac 1{1-\theta^2}\left(2\theta\sqrt R\sup_{j\geqslant 1}\left(\mathbb E\left[u_j^2\mathbf 1(A)\right]\right)^{1/2}+\sup_{j\geqslant 1}\mathbb E\left[u_j^2\mathbf 1(A)\right]\right).$$ Indeed, for each $j\geqslant 1$, \begin{align} \mathbb E\left[X_j^2\mathbf 1(A)\right]&=\theta^2\mathbb E\left[X_{j-1}^2\mathbf 1(A)\right]+2\theta\mathbb E\left[X_{j-1}u_j\mathbf 1(A)\right]+ \mathbb E\left[u_j^2\mathbf 1(A)\right]\\ &\leqslant \theta^2\sup_{j\geqslant 1}\mathbb E\left[X_{j}^2\mathbf 1(A)\right]+2\theta\sup_{j\geqslant 1}\mathbb E\left[X_{j-1}u_j\mathbf 1(A)\right]+ \sup_{j\geqslant 1}\mathbb E\left[u_j^2\mathbf 1(A)\right], \end{align} then use Cauchy-Schwarz inequality. Now, since the sequence $(u_j^2)_{j\geqslant 1}$ is uniformly integrable, we get the wanted conclusion.