$X(n)=2^{n}(1-Y(n))$ is a martingale?
Notice how $Y_{n+1}|X_0,X_1,\ldots ,X_n \sim \mathcal{U}\Big[1-2^{-n}X_n,1\Big]$ and so $$\begin{eqnarray*}\mathbb{E}\left(X_{n+1}|X_0,X_1,\dots,X_n\right) &=& \mathbb{E}\left(2^{n+1}(1-Y_{n+1})|X_0,X_1,\dots,X_n\right) \\ &=& 2^{n+1}\left(1-\mathbb{E}\left(Y_{n+1}|X_0,X_1,\dots,X_n\right)\right) \\ &=& 2^{n+1}\Big(1-\frac{1-2^{-n}X_n+1}{2}\Big) \\ &=&X_n\end{eqnarray*} $$ Hence $\{X_i\}_{i\geq 0}$ is a martingale. No integration necessary.