Relying on the below definition of uniform integrability:
Definition: A subset $\mathcal{U}$ of $\mathcal{L}^{1}$ is said to be a uniformly integrable collection of random variables if \begin{equation} \lim\limits_{c\rightarrow\infty}\sup\limits_{X\in\mathcal{U}}\mathbb{E}\{\mathbb{1}_{\{|X|\geq c\}}|X|\}=0 \end{equation}
I have to prove the following statement (Martingale Convergence Theorem):
Let $(M_n)_{n\geq1}$ be a martingale and suppose $(M_n)_{n\geq1}$ is a uniformly integrable collection of random variables. Then: \begin{equation} \lim\limits_{n\rightarrow\infty}M_n=M_{\infty} \hspace{0.5cm} \text{exists a.s.} \end{equation} $M_{\infty}$ is in $\mathcal{L}^1$ and $M_n$ converges to $M_{\infty}$ in $\mathcal{L}^1$
After proving that $\lim\limits_{n\rightarrow\infty}M_n=M_{\infty}$ exists a.s. and $M_{\infty}$ is in $\mathcal{L}^1$, I would like to prove that $M_n$ converges to $M_{\infty}$ in $\mathcal{L}^1$, that is $\lim\limits_{n\rightarrow\infty}\mathbb{E}\{|M_n - M_{\infty}|^{1}|\}=\lim\limits_{n\rightarrow\infty}\mathbb{E}\{|M_n - M_{\infty}|\}=0$.
On JACOD-PROTTER I read:
To show that $M_n$ converges to $M_{\infty}$ in $\mathcal{L}^1$, define \begin{equation} f_c (x) = \begin{cases} \hspace{0.3cm}c & \text{if} & x>c\\ \hspace{0.3cm}x & \text{if} & |x|\leq c\\ -c & \text{if} & x < -c \end{cases} \end{equation} Then, $f$ is Lipschitz (that is, there exists a real constant $c\geq0$ s.t. $\forall x_1$, $x_2 \in \mathbb{R}\hspace{0.3cm} |f_c(x_1)-f_c(x_2)| \leq c|x_1-x_2|$). BY THE UNIFORM INTEGRABILITY, THERE EXISTS $c$ SUFFICIENTLY LARGE THAT FOR $\epsilon>0$ GIVEN, IT HOLDS THAT: \begin{equation} \mathbb{E}\{|f_c(M_n)-M_n|\}<\frac{\epsilon}{3}\text{,}\hspace{0.3cm} \text{all} \hspace{0.1cm} n\text{;} \end{equation} \begin{equation} \mathbb{E}\{|f_c(M_{\infty})-M_{\infty}|\}<\frac{\epsilon}{3} \end{equation} Since $\lim M_n = M_{\infty}$ a.s., we have $\lim\limits_{n\rightarrow\infty}f_c(M_n)=f_c(M_{\infty})$, so by Lebesgue's Dominated Convergence Theorem, for $n\geq N$ ($N$ large enough): \begin{equation} \mathbb{E}\{|f_c(M_n)-f_c(M_{\infty})|\}<\frac{\epsilon}{3} \end{equation} THEREFORE, USING THE THREE ABOVE INEQUALITIES, ONE GETS: \begin{equation} \mathbb{E}\{|M_n-M_{\infty}|\}<\epsilon\text{,}\hspace{0.2cm}\text{for}\hspace{0.1cm} n\geq N \end{equation} which is equivalent to stating that $\lim\limits_{n\rightarrow\infty}\mathbb{E}\{|M_n-M_{\infty}|\}=0$, that is $M_n\rightarrow M_{\infty}$ in $\mathcal{L}^1$.
$\blacksquare$
There are two points I cannot clearly understand in JACOD-PROTTER quoted proof part:
- (FIRST PART IN BOLD CAPITAL LETTERS): Why do the first two inequalities follow from application of uniform integrability of $M_n$? I thought that they were just a consequence of the above defined Lipschitz function $f_c(x)$, such that, for a sufficiently large $c$, $f_c(M_n)=M_n$ and $f_c(M_{\infty})=M_{\infty}$. So, why are the first two inequalities a consequence of uniform integrability nature of $M_n$ and how the Lipsich function $f_c(x)$ is involved in the application of the definition of uniform integrability of a subset (defined above as well, at the beginning), which is $(M_n)$ in our case?;
- (SECOND PART IN BOLD CAPITAL LETTERS): Why if I sum both sides of the first three inequalities above, on l.h.s. I get $\mathbb{E}\{|M_n-M_{\infty}|\}$, that is why $\mathbb{E}\{|f_c(M_n)-M_n|\}+\mathbb{E}\{|f_c(M_{\infty})-M_{\infty}|\}+\mathbb{E}\{|f_c(M_n)-f_c(M_{\infty})|\}=\mathbb{E}\{|M_n-M_{\infty}|\}$?.
Re your 1st question (since the 2nd question has been already answered):
By definition, we have $f_c(x)=x$ for all $|x| \leq c$ and $f_c(x)=\pm c$ for $|x| > c$. Thus, $$|f_c(x)-x| \leq (|x|+c) 1_{|x| > c} \leq 2|x| 1_{|x| >c}.$$ Using this identity for $x=M_n(\omega)$ gives $$|f_c(M_n)-M_n| \leq 2|M_n| 1_{|M_n|>c}.$$ Taking expectation we find that $$\mathbb{E}(|f_c(M_n)-M_n|) \leq 2 \mathbb{E}(|M_n| 1_{|M_n|>c}).$$ Since $(M_n)_{n \in \mathbb{N}}$ is uniformly integrable, the right-hand side is smaller than $\frac{\epsilon}{3}$ (uniformly in $n$) for some sufficiently large constant $c=c(\epsilon)$. For $M_{\infty}$ we can use the same reasoning (recalling that $\{X\}$ is uniformly integrable for any $X \in L^1$).