If It Is a Regular Martingale, It Is a Uniformly Integrable Martingale
Definition
Let there be a given probability space $( \Omega , \mathcal{F} , P)$. When a set of random variables $\Phi$ is given, if for all $\varepsilon>0$ there exists a $k \in \mathbb{N}$ that satisfies $$ \sup_{ X \in \Phi } \int_{ \left( \left| X \right| \ge k \right) } \left| X \right| dP < \varepsilon $$, then $\Phi$ is said to be uniformly integrable. If a stochastic process $\left\{ X_{n} \right\}$ is uniformly integrable, then a martingale $\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\}$ is said to be uniformly integrable.
Theorem
If a martingale $\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\}$ is regular, it is uniformly integrable.
Explanation
Understanding why we consider $|X| \ge k$ makes it easier to accept the definition. Asking whether something is uniformly integrable in probability theory is the same as asking whether a stochastic process always possesses a first moment. In other words, it’s about checking $E |X_{n}| <\infty$, and if a natural number $k \in \mathbb{N}$ is fixed, $$ E |X_{n}| = \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP + \int_{ \left( \left| X_{n} \right| < k \right) } \left| X_{n} \right| dP $$ then $$ \begin{align*} \int_{ \left( \left| X_{n} \right| < k \right) } \left| X_{n} \right| dP <& \int_{ \left( \left| X_{n} \right| < k \right) } k dP \\ <& \int_{ \Omega } k dP \\ <& k P ( \Omega ) \\ <& \infty \end{align*} $$ and there’s no need to think about $\displaystyle \int_{ \left( \left| X_{n} \right| < k \right) } \left| X_{n} \right| dP$, only needing to check if $\displaystyle \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP$ is finite.
Proof
It suffices to show that for all $\varepsilon > 0$ there exists a $k \in \mathbb{N}$ that satisfies: $$ \sup_{ n \in \mathbb{N} } \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP < \varepsilon $$
Part 1. $\displaystyle \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP \le \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP + M {{ E |X_{n} | } \over {k}} $
Properties of conditional expectation:
- [3]: If $X$ is $\mathcal{F}$-measurable, then $E(X|\mathcal{F}) =X \text{ a.s.}$
- [10]: $\left| E( X | \mathcal{G} ) \right| \le E ( | X | | \mathcal{G} ) \text{ a.s.}$
- [11]: For all sigma fields $\mathcal{G}$, $E \left[ E ( X | \mathcal{G} ) \right] = E(X)$
Since $\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\}$ is a regular martingale, there exists an integrable random variable $\eta$ that satisfies $X_{n} = E \left( \eta | \mathcal{F}_{n} \right)$. According to the properties of conditional expectation [3] and [10], $$ \begin{align*} \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP =& \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| E \left( \eta | \mathcal{F}_{n} \right) \right| dP \\ \le & \int_{ \left( \left| X_{n} \right| \ge k \right) } E \left( | \eta| | \mathcal{F}_{n} \right) dP \\ =& \int_{ \left( \left| X_{n} \right| \ge k \right) } | \eta| dP \end{align*} $$ Now, if we split $\left( \left| X_{n} \right| \ge k \right)$ into two parts, $\left( \left| \eta \right| > M \right)$ and $\left( \left| \eta \right| \le M \right)$, $$ \begin{align*} \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP =& \int_{ \left( \left| X_{n} \right| \ge k \right) \cap \left( \left| \eta \right| > M \right) } | \eta| dP + \int_{ \left( \left| X_{n} \right| \ge k \right) \cap \left( \left| \eta \right| \le M \right)} | \eta| dP \\ \le & \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP + \int_{ \left( \left| X_{n} \right| \ge k \right) } M dP \\ \le & \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP + M P \left( \left| X_{n} \right| \ge k \right) \end{align*} $$
Markov’s inequality: $$ P(u(X) \ge c) \le {E(u(X)) \over c} $$
By Markov’s inequality, $$ \begin{align*} \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP \le & \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP + M P \left( \left| X_{n} \right| \ge k \right) \\ \le & \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP + M {{ E |X_{n} | } \over {k}} \end{align*} $$
Part 2. $\displaystyle \sup_{n \in \mathbb{N}} \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP \le \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP + {{M} \over {k}} E | \eta | \text{ a.s.}$
Since $X_{n} = E \left( \eta | \mathcal{F}_{n} \right)$, according to the properties of conditional expectation [10] and [11], $$ \begin{align*} E |X_{n} | =& E \left| E \left( \eta | \mathcal{F}_{n} \right) \right| \\ \le & E E \left( \left| \eta \right| | \mathcal{F}_{n} \right) \\ \le & E | \eta | \end{align*} $$ thus, continuing from Part 1, the following inequality is obtained. $$ \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP \le \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP + {{M} \over {k}} E | \eta | $$ This holds for all $n \in \mathbb{N}$ and $M>0$, so $$ \sup_{n \in \mathbb{N}} \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP \le \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP + {{M} \over {k}} E | \eta | $$
Part 3. $\displaystyle \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP < {{\varepsilon} \over {2}}$
Dominated convergence theorem: For a measurable set $E \in \mathcal{M}$ and $g \in \mathcal{L}^{1} (E)$, let a sequence of measurable functions $\left\{ f_{n} \right\}$ almost everywhere satisfies $|f_{n}| \le g$ in $E$. If almost everywhere in $E$ satisfies $\displaystyle f = \lim_{n \to \infty} f_{n}$, then $f \in \mathcal{L}^{1}(E)$ and $$ \lim_{ n \to \infty} \int_{E} f_{n} (x) dm = \int_{E} f dm $$
Since $|\eta| \mathbb{1}_{\left( \left| \eta \right| > M \right) } \le | \eta|$, by the dominated convergence theorem, $$ \begin{align*} \lim_{M \to \infty} \int_{ \left( \left| \eta \right| > M \right) } | \eta | dP =& \lim_{M \to \infty} \int_{ \Omega } | \eta | \mathbb{1}_{\left( \left| \eta \right| > M \right) } dP \\ =& \int_{ \Omega } \lim_{M \to \infty} | \eta | \mathbb{1}_{\left( \left| \eta \right| > M \right) } dP \\ =& 0 \end{align*} $$ In other words, for all $\displaystyle {{\varepsilon} \over {2}} > 0$, there exists a $M$ that satisfies $$ \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP < {{\varepsilon} \over {2}} $$
Part 4. $\displaystyle {{M} \over {k}} E | \eta | < {{\varepsilon} \over {2}}$
Following Part 3, there exists a $M$ that satisfies $$ \sup_{n \in \mathbb{N}} \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP \le {{\varepsilon} \over {2}} + {{M} \over {k}} E | \eta | $$ This $M$ for all $\displaystyle {{\varepsilon} \over {2}} > 0$ satisfies $$ {{M} \over {k}} E | \eta | < {{\varepsilon} \over {2}} $$ Since there exists a $k \in \mathbb{N}$ that satisfies the following for all $\varepsilon >0$, the regular martingale $\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\}$ is uniformly integrable. $$ \sup_{n \in \mathbb{N}} \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP \le {{\varepsilon} \over {2}} + {{\varepsilon} \over {2}} = \varepsilon $$
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