Random walk that is not bounded in $L^1(P)$ 
Let $T$ be a r.v. in $\mathbb{N}$ and $(Y_k)_{k\in\mathbb{N}}$ be an independent
  family of i.i.d. r.v. with $\operatorname{Var}(Y_1)=1$ and $\mathbb{E}(Y_1)=0$. Set $\mathcal{F}_n=\sigma(T,Y_1\dots,Y_n)$ and
  $$X_n:=\sum\limits_{k=1}^nY_n,\qquad n\geq 0$$
  1. Show that $(X_n)_{n\geq 0 } $ is a martingale but it is not bounded in $L^1(P)$

  
*Name a distribution of $T$ s.t. the stopped martingale $(X_{n\wedge T})_{n\in \mathbb{N}}$ is still not bounded in $L^1(P)$
  

I can show that $(X_n)$ is a martingale and I know that by the CLT
$$P\left(\frac{X_n}{\sqrt{n}}\right)\xrightarrow{n\to \infty} \mathcal{N}(0,1)$$
But I have absolutely no idea how to prove that $(X_n)$ is not bounded in $L^1(P)$. I find it strange since $\mathbb{E}(X_n)=\sum\limits_{k=1}^n\mathbb{E}(Y_n)=0$. Do I have to show somehow that it is not uniform integrable? Or is there someway to change the CLT that we can influence the expected value in the limit? 
For 2.) I also have no idea.
 A: 1 (following Kolmo's hint)
Suppose on the contrary that
$$\sup_n E[|X_n|] < \infty$$
$$\to \lim_n E[|X_n|] < \infty \ \because \ \{E[|X_n|\}_{n \in \mathbb N} \ \text{is increasing and bounded by (contrary) supposition} \ \because \ |X_n| \ \text{is a submartingale}$$
$$\to E[\liminf_n |X_n|] < \infty \ \text{by MCT}$$
$$\to \liminf_n |X_n| < \infty \ \text{a.s.}$$
$$\to |\liminf_n X_n| < \infty \ \text{a.s.}$$
$$\to \liminf_n X_n > -\infty \ \text{a.s.}$$
But
$$\liminf_n X_n = - \infty $$
QED

2
We need to find $T$ s.t.
$$\sup_n E[|X_{T \wedge n}|] = \infty$$
Let's try to visualise. If $T=3$, then
$$\sup_n E[|X_{T \wedge n}|] = \sup\{E[|X_1|], E[|X_2|], E[|X_3|], E[|X_3|], \cdots \} = \max\{E[|X_1|], E[|X_2|], E[|X_3|]\} < \infty$$
So what $T$ might make $\sup_n E[|X_{T \wedge n}|] = \infty$ ?
Let's try to follow Kolmo's hint and come up with some non-integrable $T$.
What stopping times do we know for, say, random walks (which satisfy the assumptions in your question)?
There's $T=\inf_n \{X_n = 1\}$, the the first time the sum $Y_1 + ... + Y_n$ equals 1.
Claim: $E[|T|] = E[T] = \infty$.
Pf: Suppose on the contrary that $E[|T|] = E[T] < \infty$. Then since $|X_n - X_{n-1}| = |Y_n| = 1 \le 1$, by Doob's OST, we have $$E[X_T] = E[X_1] \to E[1] = 0 \to 1 = 0 ↯$$ QED
Let's try that:
Suppose on the contrary that $$\sup_n E[|X_{T \wedge n}|] < \infty$$
$$\to \lim_n E[|X_{T \wedge n}|] < \infty$$
$$\to E[\lim_n |X_{T \wedge n}|] < \infty$$
$$\to \lim_n |X_{T \wedge n}| < \infty$$
$$\to |\lim_n X_{T \wedge n}| < \infty$$
$$\to \lim_n X_{T \wedge n} < \infty$$
Now it can be shown that $T < \infty \ \text{a.s.}$
Thus
$$\lim_n X_{T \wedge n} = X_T ( = 1)$$
$$\to 1 = X_T < \infty$$
No contradiction there, I think.
But what if we have a different stopping time $T = \infty \ \text{a.s.}$?
Then
$$\lim_n X_{T \wedge n} = \lim_n X_n$$
$$\to \lim_n X_n < \infty$$
We are finished if $\liminf_n X_n \ne \limsup_n X_n$ or $\lim_n X_n = \infty$
I think $$\limsup X_n = \infty$$ while $$\liminf X_n = -\infty$$

Also, I'm guessing there's some kind of rule that says
$$E[|T|] = E[T] < \infty \to \sup_n E[|X_{T \wedge n}|] < \infty$$
If so I think the proof is:
$$\sup_n E[|X_{T \wedge n}|] = E[|X_T|]$$
where $E[|X_T|] < \infty$ by Doob's OST.

Another approach I thought:
Since $X_n$ is a martingale, $X_{T \wedge n}$ is a martingale.
Since $X_{T \wedge n}$ is a martingale, $|X_{T \wedge n}|$ is a submartingale.
Hence, $$E[X_1] \le E[|X_{T \wedge n}|] \le E[|X_{T \wedge (n+1)}|]$$
Then since
$$\sup_n E[|X_{T \wedge n}|] = \lim_n E[|X_{T \wedge n}|]$$
if we suppose on the contrary that 
$$\sup_n E[|X_{T \wedge n}|] < \infty$$
then $\exists K > 0$ s.t.
$$E[|X_{T \wedge n}|] < K \ \forall \ n \in \ \mathbb N$$
because an increasing convergent sequence is bounded above.
Now we know that
$$E[|X_{T \wedge n}|] < \infty$$
So if we come up with some $T$ or $Y_n$ s.t. the sequence $$\{E[|X_{T \wedge n}|]\}_{n \in \ \mathbb N}$$ is not bounded above, then we're done.
