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Consider the probability space $(\Omega,\mathcal{F},\mathbb{P})$ where $\Omega=(0,1]$, $\mathcal{F}$ is the Borel $\sigma$-field generated by intervals of the form $(0,\frac{b}{2^n}]$ with $b\leq 2^n$, $b\in\mathbb{N}$, or$$ \mathcal{F}_n=\bigg\{\bigcup_j\bigg(\frac{a_j}{2^n},\frac{b_j}{2^n}\bigg]\bigg\}, $$ and $\mathbb{P}$ is the uniform Lebesgue measure. Define the random variable $Y(\omega)=\frac{1}{\omega}$.

Definition: On a general probability space $(\Omega,\mathcal{F},\mathbb{P})$, a random variable $X:\Omega\to\mathbb{R}$ is $\mathcal{F}$-measurable if $\{\omega\in\Omega:X(\omega)\leq x\}\in\mathcal{F}$ for all $x\in\mathbb{R}$.

Let $X_n$ be the largest $\mathcal{F}_n$-measurable random variable with $X_n\leq Y$. What is $X_1$ for each $\omega$?

My attempt: We have $\mathcal{F}_1=(0,1/2]\cup (1/2,1]$ and have to show that $\{\omega\in\Omega: X_1(\omega)\leq x\}\in\mathcal{F}_1$.

We also know that $X_1(\omega)\leq \frac{1}{\omega}$, where $\omega\in(0,1]$.

I struggle a lot on how do to construct such a random variable $X_1$? The part that gives me most trouble is the $\mathcal{F}_1$-measurability of $X_1$, i.e., the collection of $X_1(\omega)\leq x$ for each $x\in\mathbb{R}$ has to belong in $\mathcal{F}_1$.

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Let $B=\left(0,\frac{1}{2}\right]$. Then $\mathcal F_1=\{\varnothing, \Omega,B,B^c\}$. Clearly, $\mathcal F_1=\sigma(1_B)$. Therefore, by this theorem, $X_1$ is $f(1_B)$ for some $f:\mathbb R\to \mathbb R$ which is Borel. So, $$X_1(\omega) =\begin{cases}f(1),&\text{if }\omega\in B\\ f(0),& \text{otherwise} \end{cases}$$ So, $X_1$ is constant on $B$ and $B^c$. As $\inf_{\omega\in B} Y(\omega)=2$ (which is attained by $Y$) and $X_1\le Y$ and $X_1$ is the largest such function, $X_1(B)=\{2\}$ and similarly, $X_1(B^c)=\{1\}$. And $f(x)=x+1$ is a Borel function that satisfies $X_1=f(1_B)$.

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