Question about application of dominated convergence theorem to random variables. Prove or give a counter-example to the following statement.
It is impossible to construct a sequence of random variables $\{X_n, n ≥ 1\}$ and a random variable $X_\infty$ on the
same probability space such that the following are all true simultaneously:
• With probability 1, $X_\infty$ is a strictly positive integer.
• $E[X_\infty] = \infty.$
• For all $n\ge 1$, w.p.1, $X_n$ is an integer.
• $E[|X_n|] < \infty$ for all $n \ge 1$.
• $E[X_n] = 1$ for all $n \ge 1$.
• $\{X_n, n \ge 1\}$ converges a.s. to $X_\infty$.
I tend to prove it by using dominated convergence theorem to show $E[X_n]=1\Rightarrow E[X_\infty]<\infty$ for a contradiction, but I am wondering what dominated function I should use.
 A: This is possible, but the conditions that make it really tough to come up with a suitable counterexample are that $X_n$ is an integer, and that $E[X_n] = 1$.  If you drop the former or change the latter to $E[X_n] \in (0,2)$, then it's much more doable to come up with an example.  That being said, here's a counterexample.
Let $Y$ be the integer valued random variable with $$\mathbb{P}(Y = k) = \frac{6}{\pi^2} \frac{1}{k^2}.$$
Then $\mathbb{E}[Y] = \infty$.  Moreover, let $U$ be uniform on $[0,1]$ and $Y_1$ and $Y_2$ be indepedent copies of $Y$.  Let $\Omega$ be the probability space on which the three of them live.  The idea is that $X_k$ will be a truncation of $Y_1$ minus $Y_2$ truncated to a region quite far away from $0$.  We'll choose these correctly so that we end up with positive expectation $> 1$ but very close to $1$.  Then multiply by a suitable indicator of $U$ to get expectation $1$.  Taking $n$ to infinity will yield $X_\infty = Y_1$.
For a fixed $k$ define $Z_k = Y_1 \mathbf{1}_{Y_1 \leq k}.$  Then $\mathbb{E}{Z_k} =: c_k$ is on the order of $\log(k)$.  Define $n_k := \lceil c_k \rceil $ and set $m_k = \max\{n : c_k - \mathbb{E} (Y_2 \mathbf{1}_{k < Y_2 \leq n}) > 1 \}.$ Note that we must have $0 < c_k - \mathbb{E} (Y_2 \mathbf{1}_{k < Y_2 \leq n}) - 1 < 1/k.$  Now set $a_k = \frac{1}{c_k - \mathbb{E} (Y_2 \mathbf{1}_{k < Y_2 \leq n})}$.  We can now define: $$X_k = \mathbf{1}_{U < a_k}\cdot\left(Y_1 \cdot \mathbf{1}_{Y_1 \leq k} - Y_2 \cdot \mathbf{1}_{k < Y_2 \leq n_k} \right).$$
One can check that $X_n \to Y_1$ almost surely (this follows from the fact that $a_k \to 1$), implying the first two and last condition.  By construction $X_n$ is an integer.  We know that $|X_n|$ is bounded, so $\mathbb{E}|X_n| <\infty$.  By independent, $\mathbb{E}[X_n] = 1$.
