# Why does $\mathbb{E}(F(X) 1_A) = \mathbb{P}(A) \mathbb{E}(F(X))$ imply the independence of $X$ and $A$?

I was trying to understand the Markov property of Brownian Motion and in one of the proofs the author claims that the following result

Let $\mathcal{F} \subseteq \mathcal{A}$ be a $\sigma$-algebra and $X:\Omega \to \mathbb{R}^d$ be a random variable such that $$\mathbb{E}(F(X) 1_A) = \mathbb{P}(A) \mathbb{E}(F(X))$$ for all $A \in \mathcal{F}$ and $F$ bounded, continuous. Then $X$ and $\mathcal{F}$ are independent.

I understand that if we know that they were independent then the above equality will be true but i cannot prove why this implies independence if I use the basic definition of independence . Can you give me some hints on how could I go about showing it ?I have thought a lot about it ? But I was unsuccessful in proving it .

Suppose that

$$\mathbb{E}(F(X) 1_A) = \mathbb{P}(A) \mathbb{E}(F(X)) \qquad \text{for all A \in \mathcal{F}}. \tag{1}$$

Since

$$e^{i \eta 1_A} = e^{i \eta} 1_{A} + 1_{A^c}\tag{2}$$

for any $\eta \in \mathbb{R}$ we find \begin{align*} \mathbb{E}(e^{i \xi X} e^{i \eta 1_A}) &\stackrel{(2)}{=} e^{i \eta} \mathbb{E}(e^{i \xi X} 1_A) + \mathbb{E}(e^{i \xi X} 1_{A^c}) \\ &\stackrel{(1)}{=} e^{i \eta} \mathbb{P}(A) \mathbb{E}(e^{i \xi X}) + \mathbb{P}(A^c) \mathbb{E}(e^{i \xi X}) \\ &\stackrel{(2)}{=} \mathbb{E}(e^{i \eta 1_A}) \mathbb{E}(e^{i \xi X}) . \end{align*}

This implies that $X$ and $1_A$ are independent (see this answer for details).

Alternative proof: For any closed set $G$ there exists a sequence of bounded continuous functions $(F_n)_{n \in \mathbb{N}}$ such that $F_n \downarrow 1_G$. Using the monotone convergence theorem, we get

$$\mathbb{E}(1_G(X) 1_A) = \mathbb{P}(A) \mathbb{E}(1_G(X)).$$

Since the closed sets generate the Borel-$\sigma$-algebra, it is not difficult to see that this implies

$$\mathbb{E}(1_B(X) 1_A) = \mathbb{P}(A) \mathbb{E}(1_B(X)) \quad \text{for all B \in \mathcal{B}(\mathbb{R})^d), A \in \mathcal{F}}$$

which shows that $X$ and $\mathcal{F}$ are independent.

• Pretty neat!$\,$ – Stefan Hansen Feb 3 '17 at 6:29
• Very nice work, but I can't see why your 1st claim in the alternative proof is true. Can you briefly explain to me why it is true and how to construct the $F_n$ or give me a reference that explains the claim? Thanks a lot – Ansel B Feb 3 '17 at 12:14
• @AnselB Define $$F_n(x) := \max\{0,1-n \cdot d(x,G)\}$$ where $$d(x,G) := \inf_{y \in G} |x-y|.$$ – saz Feb 3 '17 at 12:26
• Thanks, I am guessing that this only work for closed sets as if G wasn't closed then $F_n\downarrow 1_{\bar{G}}$? Is there a function that will work for any set? – Ansel B Feb 3 '17 at 13:31
• @AnselB If $G$ is not closed, then there does, in general, not exist such a sequence decreasing to $1_G$. For instance if $G$ is open it is possible to construct continuous bounded $F_n \geq 0$ such that $F_n \uparrow 1_G$ but there doesn't exist a sequence $F_n$ such that $F_n \downarrow 1_G$. (If you are interested in this, search for "Urysohn's lemma".) – saz Feb 3 '17 at 13:35