Sequence which converges in probability but neither almost surely nor in $L^p$ Is it possible to find a sequence of r.v.'s $\{X_n\}$ that converge in probability but not almost surely nor in $L^p$?
Does anyone know of a single example that illustrates this point? I've been trying to think of one for ages and I've not come up with anything. I'm sure such a sequence must exist, but can't think what it could be.
 A: Perhaps a more probabilistic example:
For each $n$, let $Z_n$ be uniformly distributed on the set $\{1, 2, \dots, n\}$.  For $1 \le k \le n$, let
$$X_{n,k} = \begin{cases} n^2, & \text{if $Z_n = k$}\\ 0, & \text{otherwise}.\end{cases}$$
Then consider the sequence $\{X_{1,1}, X_{2,1}, X_{2,2}, \dots, X_{n,1}, \dots, X_{n,n}, \dots\}$ which steps through the "rows" of this triangular array.  It converges to 0 in probability since for any $\epsilon &lt 1$, we have $P(|X_{n,k}| > \epsilon) = P(Z_n = k) = 1/n$ which goes to $0$ as $n \to \infty$.  It does not converge almost surely, since the sequence will always have infinitely many values greater than 1 (one in each row) but all the rest are zero.  Finally, we have $E|X_{n,k}|^p = \frac{1}{n} n^{2p} = n^{2p-1} \to \infty$ for all $p \ge 1$, so the sequence does not converge in $L^p$ either.
A: $X_n$ independent such that $P(X_n=n)=\frac{1}{n}$ and $P(X_n=0)=1-\frac{1}{n}$.
Then $\forall \varepsilon>0$, $P(|X_n|>\varepsilon)=\frac{1}{n} \to 0$ but by second Borel Cantelli $P(\limsup |X_n|>\varepsilon)=1$ and $E|X_n|^p=n^{p-1}$ that doesn't converge to 0 $\forall p\geq1$.
A: Let $P$ be the Lebesgue measure on $[0,1]$ and
$$f_{n,k}(x) = \begin{cases} 4^n & x\in \left[\frac{k-1}{2^n}, \frac{k}{2^n}\right] \\ 0 & \text{otherwise} \end{cases}$$
for $k\in \{1, \dots, 2^n\}$. 
Then $$\|f_{n,k}\|_1 =\frac{4^{n}}{2^{n}} = 2^n$$ $f_{n,k}$ doesn't converge pointwise (going through the sequence $(n,k)$ as $(1,1), \;(2,1),\; (2,2),\; (3,1),\;   \dots$) and $$P[f_{n,k} > \epsilon] = 2^{-n}\to 0$$ for all $0&lt\epsilon&lt1$.
