Proof of Lévy's Theorem in Probability Theory
Theorem
Let’s assume that a probability space $( \Omega , \mathcal{F} , P)$ is given.
If $\eta$ is an integrable random variable and $\left\{ \mathcal{F}_{n} \right\}_{n \in \mathbb{N}}$ is a sequence of sigma fields where $\left\{ \mathcal{F}_{n} \right\}_{n \in \mathbb{N}}$ is $\mathcal{F}_{n} \subset \mathcal{F}_{n+1}$, then $n \to \infty$ when $$ E \left( \eta | \mathcal{F}_{n} \right) \to E \left( \eta | \mathcal{F}_{\infty} \right) $$
- $\displaystyle \mathcal{F}_{\infty} = \bigotimes_{n=1}^{\infty} \mathcal{F}_{n}$ does not mean the tensor product but represents the smallest sigma field containing all elements of $\mathcal{F}_{n}$. It’s not particularly new, as the smallest sigma field containing all open sets of a topological space $\Omega$ has been referred to as the Borel sigma field. However, if it’s difficult, it could simply be taken as a sigma field satisfying the filtration condition.
Explanation
Unlike the Lebesgue’s theorem in measure theory, where the integrand remains static and the sigma field expands, the essence is not significantly different.
Lévy’s 0-1 Law
Lévy’s theorem, also known as Lévy’s zero–one law, suggests that the conditional probability $P \left( A | \mathcal{F}_{\infty} \right) = E \left( 1_{A} | \mathcal{F}_{\infty} \right)$ of an event $A \in \mathcal{F}_{\infty}$ is almost surely $$ P \left( A | \mathcal{F}_{n} \right) \to 1_{A} \in \left\{ 0 , 1 \right\} $$ That is, either $0$ or $1$ when $n \to \infty$. Intuitively, the fact that the sigma field expands while satisfying $\mathcal{F}_{n} \subset \mathcal{F}_{n+1}$, resulting in the filtration of sigma fields, means an increase in the amount of information, which clarifies whether the event $A$ occurs as either $0$ or $1$1.
Proof
Strategy: It’s necessary to employ the properties of regular martingales in probability theory and Doob’s martingale convergence theorem. Note that since $\eta$ is only given as an integrable random variable, it cannot assert that it’s almost surely $X_{\infty} = \eta$ for all $A \in \mathcal{F}_{\infty}$ as it’s not $\mathcal{F}_{\infty}$-measurable. However, according to the properties of conditional expectation, $E \left( \eta | \mathcal{F}_{\infty} \right)$ is $\mathcal{F}_{\infty}$-measurable regardless of what $\eta$ is, and the equation to be proved actually becomes $X_{\infty} = E \left( \eta | \mathcal{F}_{n} \right)$.
Claim: Let’s define $X_{n} : = E \left( \eta | \mathcal{F}_{n} \right)$ and a random variable $\displaystyle X_{\infty} := \lim_{n \to \infty} X_{n}$ that is $\mathcal{F}_{\infty}$-measurable. It’s necessary to prove $X_{\infty} = E \left( \eta | \mathcal{F}_{\infty} \right)$.
Part 1. $E(X_{\infty} | \mathcal{F}_{n}) = X_{n}$
According to the definition of $X_{n}$, $\left\{ (X_{n} , \mathcal{F}_{n} ) \right\}$ is a regular martingale. Therefore, it’s a uniformly integrable martingale, and becomes a $\mathcal{L}_{1}$ convergent martingale, converging to $X_{\infty}$ in the sense of $\mathcal{L}_{1}$. Furthermore, since $\left\{ (X_{n} , \mathcal{F}_{n} ) \right\}$ is a closable martingale, $E(X_{\infty} | \mathcal{F}_{n}) = X_{n}$ is obtained.
Now, to use Doob’s martingale convergence theorem, the following definitions are introduced.
- A $\mathcal{P}$ satisfying the following is called a $\pi$-system. $$ A, B \in \mathcal{P} \implies A \cap B \in \mathcal{P} $$
- A $\mathcal{L}$ satisfying the following conditions is called a $\lambda$-system.
- (i): $\emptyset \in \mathcal{L}$
- (ii) $A \in \mathcal{L} \implies A^{c} \in \mathcal{L}$
- (iii) For all $i \ne j$, when $\displaystyle A_{i} \cap A_{j} = \emptyset$ $$\displaystyle \left\{ A_{n} \right\}_{n \in \mathbb{N}} \subset \mathcal{L} \implies \bigcup_{n \in \mathbb{N}} A_{n} \in \mathcal{L}$$
Part 2. $\displaystyle \mathcal{L} := \left\{ A \in \mathcal{F} : \int_{A} X_{\infty} dP = \int_{A} \eta dP \right\}$ is a Lambda system
Part 2-(i). $\emptyset \in \mathcal{L}$
- By the empty set axioms, $\emptyset \in \mathcal{L}$ is trivial.
Part 2-(ii). $A \in \mathcal{L} \implies A^{c} \in \mathcal{L}$
- As Part 1 implied $E(X_{\infty} | \mathcal{F}_{n}) = X_{n}$, according to the definition of conditional expectation, $$ \begin{align*} \int_{\Omega} X_{\infty} dP =& \int_{\Omega} E ( X_{\infty} | \mathcal{F}_{n} ) dP \\ =& \int_{\Omega} X_{n} dP \\ =& \int_{\Omega} E \left( \eta | \mathcal{F}_{n} \right) dP \\ =& \int_{\Omega} \eta dP \end{align*} $$ thus $\Omega \in \mathcal{L}$, and by the definition of $\mathcal{L}$, if $A \in \mathcal{L}$, $$ \begin{align*} \int_{A^{c}} X_{\infty} dP =& \int_{\Omega} X_{\infty} dP - \int_{A} X_{\infty} dP \\ =& \int_{\Omega} \eta dP - \int_{A} \eta dP \\ =& \int_{A^{c}} \eta dP \end{align*} $$ This calculation was possible because the probability measure $P$ excludes cases like $\infty - \infty$ being a finite measure. Therefore, by the calculation $A^{c} \in \mathcal{L}$, and in conclusion, $$ A \in \mathcal{L} \implies A^{c} \in \mathcal{L} $$
Part 2-(iii). For all $i \ne j$, when $\displaystyle A_{i} \cap A_{j} = \emptyset$ $\displaystyle \left\{ A_{n} \right\}_{n \in \mathbb{N}} \subset \mathcal{L} \implies \bigcup_{n \in \mathbb{N}} A_{n} \in \mathcal{L}$
- $$\begin{align*} \int_{\bigcup_{i \in \mathbb{N}} A_{i}} X_{\infty} dP =& \sum_{i=1}^{\infty} \int_{A_{i}} X_{\infty} dP \\ =& \sum_{i=1}^{\infty} \int_{A_{i}} \eta dP \\ =& \int_{\bigcup_{i \in \mathbb{N}} A_{i}} \eta dP \end{align*}$$
Part 3. $\displaystyle \mathcal{P} := \bigcup_{n \in \mathbb{N}} \mathcal{F}_{n}$ is a Pi system
If $\displaystyle A, B \in \bigcup_{n \in \mathbb{N}} \mathcal{F}_{n}$, then for some $n_{1}, n_{2} \in \mathbb{N}$, $$ A \in \mathcal{F}_{n_{1}} \\ B \in \mathcal{F}_{n_{2}} $$ Therefore, $$ (A \cap B) \in \mathcal{F}_{\max \left\{ n_{1} , n_{2} \right\} } \subset \bigcup_{n \in \mathbb{N}} \mathcal{F}_{n} $$
Part 4. $\displaystyle \mathcal{P} \subset \mathcal{L}$
To say that $\displaystyle A \in \mathcal{P} = \bigcup_{n \in \mathbb{N}} \mathcal{F}_{n}$ is to suggest the existence of some $n_{0} \in \mathbb{N}$ that satisfies $A \in \mathcal{F}_{n_{0}}$. Hence, since Part 1 showed $\left\{ (X_{n} , \mathcal{F}_{n} ) \right\}$ was a martingale, $$ \begin{align*} \int_{A} X_{m} dP =& \int_{A} E ( X_{m}| \mathcal{F}_{n_{0}} ) dP \\ =& \int_{A} X_{n_{0}} dP \\ =& \int_{A} E \left( \eta | \mathcal{F}_{n_{0}} \right) dP \\ =& \int_{A} \eta dP \end{align*} $$ Therefore, $$ \begin{align*} \left| \int_{A} X_{m} dP - \int_{A} X_{\infty} dP \right| \le & \int_{A} \left| X_{m} - X_{\infty} \right| dP \\ \le & E | X_{m} - X_{\infty} | \end{align*} $$ Here, since $\left\{ (X_{n} , \mathcal{F}_{n} ) \right\}$ is a martingale converging in the sense of $\mathcal{L}_{1}$, when $m \to \infty$, $E| X_{m} - X_{\infty}| \to 0$ and $$ \lim_{m \to \infty} \int_{A} X_{m} dP = \int_{A} X_{\infty} dP $$ And for all $m \in \mathbb{N}$, $$ \int_{A} X_{m} dP = \int_{A} E \left( \eta | \mathcal{F}_{m} \right) dP = \int_{A} \eta dP $$ Therefore, $$ \begin{align*} \int_{A} X_{\infty} dP =& \lim_{m \to \infty} \int_{A} X_{m} dP \\ =& \lim_{m \to \infty} \int_{A} \eta dP \\ =& \int_{A} \eta dP \end{align*} $$ Thus, by the definition of $\mathcal{L}$, $A \in \mathcal{L}$. In conclusion, $$ A \in \mathcal{P} \implies A \in \mathcal{L} \\ \mathcal{P} \subset \mathcal{L} $$
Part 5.
Doob’s martingale convergence theorem: If a Pi system $\mathcal{P}$ is a subset of a Lambda system $\mathcal{L}$, there exists a sigma field $\sigma ( \mathcal{P} )$ satisfying $\mathcal{P} \subset \sigma ( \mathcal{P} ) \subset \mathcal{L}$.
$$ \begin{align*} \sigma \left( \mathcal{P} \right) =& \sigma \left( \bigcup_{n \in \mathbb{N}} \mathcal{F}_{n} \right) \\ =& \bigotimes_{n=1}^{\infty} \mathcal{F}_{n} \\ =& \mathcal{F}_{\infty} \end{align*} $$ According to Doob’s martingale convergence theorem, there exists a sigma field $\mathcal{F}_{\infty}$ satisfying the following: $$ A \in \mathcal{F}_{\infty} \implies \int_{A} X_{\infty} d P = \int_{A} \eta dP $$ Therefore, for all $A \in \mathcal{F}_{\infty}$, $$ \int_{A} X_{\infty} d P = \int_{A} \eta dP = \int_{A} E \left( \eta | \mathcal{F}_{\infty} \right) dP $$ Since $X_{\infty}$ and $E \left( \eta | \mathcal{F}_{\infty} \right)$ are $\mathcal{F}_{\infty}$-measurable, according to the properties of the Lebesgue integral, almost surely $X_{\infty} = E \left( \eta | \mathcal{F}_{\infty} \right)$. Then, as initially defined, $X_{n}$ and $X_{\infty}$, $$ \begin{align*} \lim_{n \to \infty} E \left( \eta | \mathcal{F}_{n} \right) =& \lim_{n \to \infty} X_{n} \\ =& X_{\infty} \\ =& E \left( \eta | \mathcal{F}_{\infty} \right) \end{align*} $$
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