# Prove $X_n=2^n\textbf{1}_{(0,2^{-n}]}$ is a martingale

Let $$((0,1], \text{Borel}_{(0,1]}, \mu)$$ be a measure space, where $$\mu$$ is the Lebesgue measure. I want to prove that $$\{X_n\}_{n\geq 0}$$ is a martingale with respect to the filtration $$\mathcal{F}_n=\sigma(X_1,\ldots,X_n)$$, where $$X_n=2^n\textbf{1}_{(0,2^{-n}]}$$. I only need help to prove the third property of martingales, that is, $$\mathbb{E}(X_n\mid\mathcal{F}_{n-1})=X_{n-1}$$ $$\forall n\geq1$$.

So far this is what I've tried:

\begin{align} \mathbb{E}(X_n\mid\mathcal{F}_{n-1})&= \mathbb{E}(2^n\textbf{1}_{(0,2^{-n}]}\mid\mathcal{F}_{n-1}) \\&=2\mathbb{E}(2^{n-1}\textbf{1}_{(0,2^{-n}]}\mid\mathcal{F}_{n-1}) \\&=2\mathbb{E}[2^{n-1}(\textbf{1}_{(0,2^{-n+1}]}-\textbf{1}_{(2^{-n},2^{-n+1}]})\mid\mathcal{F}_{n-1}] \\&=2X_{n-1}-\mathbb{E}[2^n\textbf{1}_{(2^{-n},2^{-n+1}]}\mid\mathcal{F}_{n-1}] \end{align}

If I were working with normal expectantions instead of conditional ones, I could write: $$\mathbb{E}[2^n\textbf{1}_{(2^{-n},2^{-n+1}]}\mid\mathcal{F}_{n-1}]=\mathbb{E}[2^n\textbf{1}_{(0,2^{-n}]}\mid\mathcal{F}_{n-1}]$$ because $$(0, 2^{-n}]$$ and $$(2^{-n},2^{-n+1}]$$ have the same probability under Lebesgue measure, and then my problem would be solved because I would get: $$\mathbb{E}(X_n\mid\mathcal{F}_{n-1})=2X_{n-1}-\mathbb{E}(X_n\mid\mathcal{F}_{n-1}) \Longrightarrow \mathbb{E}(X_n\mid\mathcal{F}_{n-1})=X_{n-1}$$ However I don't think I can use this technique here, because:

$$\exists A\in \mathcal{F}_{n-1} \quad \mbox{so that} \quad \int_{A}\textbf{1}_{(0,2^{-n}]}d\mu \neq \int_{A}\textbf{1}_{(2^{-n}, 2^{-n+1}]}d\mu$$ And therefore the conditional expectations are not the same.

Since $$X_n$$ is $$\newcommand{\F}{\mathcal{F}}\F_n$$-measurable you only need to prove that $$\newcommand{\E}{\mathbf{E}} \forall A\in\F_n,\quad\E[X_{n+1}1_A] \overset{(\star)}= \E[X_{n}1_A].$$ Furthermore, every measurable $$A\in\F_n$$ is a disjoint union of intervals from the set $$\Big\{\big(0,2^{-n}\big], ~\big(2^{-n}, 2^{-n+1}\big], ~\dots~,~ \big(1/4,1/2\big],~ \big(1/2, 1\big]\Big\}$$ and so you only need to prove $$(\star)$$ for $$A$$'s chosen from this set of intervals. All equalities are clear (they are all $$0=0$$ except the first one which is $$1=1$$) so that $$X_n=\E[X_{n+1}\mid\F_{n}]$$