$\{\mathbb {E}^{\mathcal {B}}X~:~\mathcal {B}$ is a sub-$\sigma$-algebra of $\mathcal{F} \}$ is uniformly integrable

Question

Let $X\in L^{1}(\Omega , \mathcal{F},\mathbb {P}) $. Show that :

$$ H=\{\mathbb {E}^{\mathcal {B}}X~:~\mathcal {B}~\text{is a sub-$\sigma$-algebra of $\mathcal{F}$}\} $$ is uniformly integrable (UI).

My effort:

$1. $ $\forall \mathcal {B}\subset \mathcal{F} $: $$ \int_{\Omega} |\mathbb {E}^{\mathcal {B}}X|d\mathbb {P}\leq \int_{\Omega} \mathbb {E}^{\mathcal {B}}|X|d\mathbb {P}= \int_{\Omega}|X|d\mathbb {P}<M~~(\text{because}~ X\in L^1) $$ Then :

$$ \sup_{\mathcal {B}\subset \mathcal{F} }\int_{\Omega} |\mathbb {E}^{\mathcal {B}}X|d\mathbb{P}<+\infty $$ $2. $ my problem is to show that :

$\forall \epsilon >0, \exists \sigma >0$ such as:

$$ [\forall A\in \mathcal{F}\, \text{such as:}\, \mathbb {P}(A)<\sigma ] \Rightarrow [\forall \mathcal {B}\subset \mathcal{F} ~:~ \int_{A} |\mathbb {E}^{\mathcal {B}}X|d\mathbb {P}<\epsilon ] $$

Answer 1

Fix $\epsilon>0$. For any $A \in \mathcal{F}$ and $R>0$, we have

\begin{align*} \int_A |\mathbb{E}^{\mathcal{B}}X| \, d\mathbb{P} &= \int_{A \cap \{|\mathbb{E}^{\mathcal{B}}X| \leq R\}} |\mathbb{E}^{\mathcal{B}}X| \, d\mathbb{P} + \int_{A \cap \{|\mathbb{E}^{\mathcal{B}}X|>R\}} |\mathbb{E}^{\mathcal{B}}X| \, d\mathbb{P} \\ &\leq R \mathbb{P}(A) + \int_{\{|\mathbb{E}^{\mathcal{B}}X|>R\}} \mathbb{E}^{\mathcal{B}}(|X|) \, d\mathbb{P}. \tag{1} \end{align*}

As $\{|\mathbb{E}^{\mathcal{B}}X|>R\} \in \mathcal{B}$, it follows from the definition of the conditional expectation that

$$\int_{\{|\mathbb{E}^{\mathcal{B}}X|>R\}} \mathbb{E}^{\mathcal{B}}(|X|) \, d\mathbb{P} = \int_{\{|\mathbb{E}^{\mathcal{B}}X|>R\}} |X| \, d\mathbb{P}.$$

Now

\begin{align*} \int_{\{|\mathbb{E}^{\mathcal{B}}X|>R\}} \mathbb{E}^{\mathcal{B}}(|X|) \, d\mathbb{P} &= \int_{\{|X| \leq K\} \cap \{|\mathbb{E}^{\mathcal{B}}X|>R\}} |X| \, d\mathbb{P} + \int_{\{|X|>K\} \cap \{|\mathbb{E}^{\mathcal{B}}X|>R\}} |X| \, d\mathbb{P} \\ &\leq K \mathbb{P}(|\mathbb{E}^{\mathcal{B}}X|>R) +\int_{\{|X|>K\}} |X| \, d\mathbb{P} \\ &\leq \frac{K}{R} \underbrace{\mathbb{E}(|\mathbb{E}^{\mathcal{B}}X|)}_{\leq \mathbb{E}(|X|)} +\int_{\{|X|>K\}} |X| \, d\mathbb{P} . \end{align*}

Plugging this into $(1)$ yields

$$ \int_A |\mathbb{E}^{\mathcal{B}}X| \, d\mathbb{P} \leq R \mathbb{P}(A) + \frac{K}{R} \mathbb{E}(|X|) + \int_{\{|X|>K\}} |X| \, d\mathbb{P}$$

for all $K,R>0$ and $A \in \mathcal{F}$. Now choose

• $K>0$ sufficiently large such that $\int_{|X|>K} |X| \, d\mathbb{P} < \epsilon/3$,
• $R>0$ sufficiently large such that $\frac{K}{R} \mathbb{E}(|X|) \leq \epsilon/3$ (for the $K$ which we have already chosen)

Set $\sigma:=\epsilon/(3R)$ for $R$ which we have just chosen. Then

$$ \int_A |\mathbb{E}^{\mathcal{B}}X| \, d\mathbb{P} <\epsilon$$

for all $A \in \mathcal{F}$ with $\mathbb{P}(A)<\sigma$ and for all sub-$\sigma$-algebras $\mathcal{B} $.