A martingale that does not converge in $L^1$ 
I am currently studying martingales and I am working on the following problem:
Let $\Omega = \mathbb N^*$ and associated probability measure
   $$\forall\, k \in \mathbb N^*, P({\{k\}})=\frac{1}{k}-\frac{1}{k+1}$$
  For $\,n \ge 1$, let the $X_n=(n+1)\,\mathbf{1}_{[n+1,\, \infty[}\,,$ and $\mathscr F_n=\sigma(\{1\},\{2\},\dots, \{n\},[n+1,\, \infty[)$
Show that:
  
  
*
  
*$(X_n)_{n \ge 0}$ is a martingale with respect to $\mathscr{(F_n)_{n \ge 0}}$
  
*The $(X_n)_{n \ge 0}$ sequence converges $a.s.$ but not in $\mathbb{L}^1(\Omega)$. Is it uniformly integrable?
  



*

*Martingale


I would like to prove that $\mathbb{E}[X_n\mid\mathscr{F_{n-1}}]=X_{n-1}$
First of all every $X_n$ is $\mathbb{L}^{\infty}(dP)$ hence $\mathbb{L}^1(dP)$, and is $\mathscr{F}_n$-measurable.
We note that we can write $F_{n-1}$ as a partition consisting of the $n$ elements $\{\{1\},\{2\},\dots, \{n-1\},[n,\, \infty[\}$.
We can therefore use the formula for the conditional expectation with respect to a σ-algebra generated by that partition.
$$\begin{align*} \mathbb{E}[X_n\mid\mathscr{F_{n-1}}] &= \mathbb{E}\big[\mathbf{1}_{[n,\infty]}\cdot X_n \mid [n,\infty]\big]+\sum_{k=1}^{n-1}\mathbb{E}\big[\mathbf{1}_{\{k\}}\cdot X_n \mid \{k\}\big] 
\\ &= \mathbb{E}\big[\mathbf{1}_{[n,\infty[}\cdot (n+1)\, \mathbf{1}_{[n+1,\infty[} \mid [n,\infty[\big] + \sum_{k=1}^{n-1}\mathbb{E}\big[\mathbf{1}_{\{k\}}\cdot (n+1)\, \mathbf{1}_{[n+1,\infty[} \mid \{k\}\big] 
\\ &= (n+1)\,\mathbf{1}_{[n,\infty[}\cdot \frac{\mathbb{E}\big[\mathbf{1}_{[n+1,\infty[}\, \mathbf{1}_{[n,\infty[}\big]} {\mathbb{P}([n,\infty[)} + (n+1)\,\sum_{k=1}^{n-1} \mathbf{1}_{\{k\}}\cdot \frac{\mathbb{E}\big[ \mathbf{1}_{[n+1,\infty[}\,\mathbf{1}_{\{k\}} \big]}{\mathbb{P}(\{k\})} 
\\ &= (n+1)\,\mathbf{1}_{[n,\infty[} \cdot\frac{\mathbb{P}({[n+1,\infty[\, \cap}\,{[n,\infty[})} {\mathbb{P}([n,\infty[)} + (n+1)\,\sum_{k=1}^{n-1} \mathbf{1}_{\{k\}}\cdot \frac{\mathbb{P}({[n+1,\infty[\, \cap}\,{\{k\}})}{\mathbb{P}({\{k\}})}\end{align*}$$
The summation on the right-hand side gives $0$ since ${[n+1,\infty[\, \cap}\,{\{k\}}=\emptyset, \ \forall k\, \in [1, n-1]$. So we are left with:
$$\begin{align*} \mathbb{E}[X_n\mid\mathscr{F_{n-1}}]&= (n+1)\,\mathbf{1}_{[n,\infty[}\cdot \frac{\mathbb{P}({[n+1,\infty[\,})} {\mathbb{P}([n,\infty[)}
\end{align*}$$
We have the telescopic sum: $$ \mathbb{P}({[n,\infty[})= \sum_{k=n}^{\infty}\Big(\frac{1}{k}-\frac{1}{k+1}\Big) = \frac{1}{n}$$
Hence:
$$\begin{align*} \mathbb{E}[X_n\mid\mathscr{F_{n-1}}]&= (n+1)\,\mathbf{1}_{[n,\infty[}\cdot \frac{1}{n+1}\cdot n = n \, \mathbf{1}_{[n,\infty[} 
\\ &= X_{n-1} \end{align*}$$


*Convergence


*

*Almost sure convergence



My guess is that $X_n \xrightarrow[\infty]{a.s.}{0}$, so I want to prove that $$\forall\, \epsilon \ge 0, \lim_{{n}\to {\infty}} \mathbb{P}\,(\sup_{k\ge n} |X_n| \ge \epsilon) = \lim_{{n}\to {\infty}} \mathbb{P}(\sup_{k\ge n}\, (k+1)\,\mathbf{1}_{[k+1, +\infty]} \ge \epsilon)=0$$
$$\begin{align*} \lim_{{n}\to {\infty}} \mathbb{P}(\sup_{k\ge n}\, (k+1)\,\mathbf{1}_{[k+1, +\infty]} \ge \epsilon) &= \lim_{{n}\to {\infty}} \mathbb{P}({[n+1, +\infty]}) \\ &= \lim_{{n}\to {\infty} }\frac {1}{n+1} = 0 \end{align*}$$
Thus, for every $\omega \in \Omega$, for every large enough $n$, $X_n(\omega)=0$. Therefore $(X_n)_{n \ge 0}$ converges $a.s.$ to $0$. 


*

*Convergence in $\mathbb{L}^1(\Omega)$
I would like to show that
$$\mathbb{E}(|X_n|) \xrightarrow[n\to\infty]{}{0} $$
We have:
$$\begin{align*} \mathbb{E}(|X_n|) &= \mathbb{E}\big[(n+1)\,\mathbf{1}_{[n+1, +\infty]}\big] = (n+1)\,\mathbb{E}\Big[\sum_{k=n+1}^{\infty}\mathbf{1}_{\{k\}}\Big] 
\\ &= (n+1)\,\sum_{k=n+1}^{\infty}\mathbb{E}(\mathbf{1}_{\{k\}}) = (n+1)\,\sum_{k=n+1}^{\infty} \Big( \frac{1}{k}-\frac{1}{k+1} \Big) 
\\ &= (n+1)\cdot \frac{1}{n+1} = 1 \neq 0 \end{align*} $$
showing that $(X_n)_{n \ge 0}$ does not converge in $\mathbb{L}^1(\Omega)$.


*

*$(X_n)_{n \ge 0}$ is uniformly integrable


According to 1 (item 6) convergence in $\mathbb{L}^1$ is equivalent to uniform integrability combined with convergence in probability. Since $(X_n)$ does not converge in $\mathbb{L}^1$, either one or both requirements are wrong. 
But we know $(X_n)$ converges $a.s.$, which implies convergence in probability. Therefore it must be the uniform integrability requirement which is not met.
We conclude $(X_n)$ is not uniformly integrable.
I have edited my initial solution based on the comments. 
 A: *

*You forgot to point out that every $X_n$ is $L^{\infty}(dP)$ hence $L^1(dP)$, and is $\mathscr{F}_n$-measurable.


I am also a bit puzzled over your « partition formula » in your computation of the conditional expectation. Can you find some reference or more basic proof or explain?


*Almost sure convergence 


I am not sure your criterion is enough: you are basically claiming that if $(Y_n)$ is a decreasing sequence of nonnegative rv, then if $Y_n$ converges to $0$ in probability it converges to $0$ almost surely.
I would suggest the following proof: for every $\omega \in \Omega$, for every large enough $n$ $X_n(\omega)=0$. Thus $X_n$ converges as to $0$. 


*$L^1$ convergence: the expected value of every $X_n$ is $1$, (you forgot to take the indicatir function inside $X_n$ into account). You could have noticed the mistake by seeing that the expected value of terms of a martingale has to be constant.


However you cannot have $L^1$ convergence of any subsequence to any random variable $Y$ because then $Y$ is non-negative with expectation $1$; however since $X_n$ converges to $0$ in law, $Y$ must be $0$, a contradiction. 
(Note that from Markov inequality $L^1$ convergence implies convergence in probability implies convergence in law). 
Uniform integrability: note that over $\mathbb{N^*}$ and any probability, a subspace of $L^1$ is compact iff it is bounded and uniformly integrable. Since no subsequence of the $X_n$ converges in $L^1$, and the sequence is bounded in $L^1$, no infinite subset of them is uniformly integrable.
A: Here are some remarks.


*

*You write $\mathcal F_{n-1}$ as a partition consisting of two elements, but actually we have to write it as a partition consisting of $n$ elements and use the formula for the conditional expectation with respect to a $\sigma$-algebra generated by a partition, which you did. By the way, it seems that you do not really need $G_n$, $G_1$ and $G_2$.

*For the convergence in $\mathbb L^1$, there is a mistake in the computation of $\mathbb P\left(\left[n+1,+\infty\right]\right)$, while you did it correctly in the previous step.

*As a consequence of this mistake, the justification of the uniform integrability is not correct (the expectation of the absolute value is one). The fact that there is no convergence in $\mathbb L^1$ combined with item 6 of the linked answer gives a justification.

