I'm currently working on the following problem: Construct a sequence of random variables $X_n$ and $X$ such that $X_n \overset{\mathbb{P}}{\rightarrow} X$ and $\mathbb{E}[X_n] = 0$ for all $n$ but $\mathbb{E}[X] = \infty$.
I can imagine random variables with infinite expectation like $\mathbb{P}[X] = \frac{1}{n(n+1)}$ for all $n \in \mathbb{N}$ but struggle to see how to construct a sequence that converges to $X$ while meeting the criteria.
On
Take for example: $$ X_n = \begin{cases} X & X<n \\ -n\ln n & X\geq n \end{cases} $$ with $X$ a Pareto distribution with exponent 1, i.e. it is supported on $(1,+\infty)$ with pdf: $$ p(x) = \frac1{x^2} $$ Using: $$ \mathbb P(X>n) = \int_n^{+\infty}\frac{dx}{x^2} = \frac1n $$ You can check that: $$ \begin{align} \mathbb E(X_n) &= \int_1^n \frac{xdx}{x^2}-n\ln n\frac1n\\ &= \ln n-\ln n \\ &= 0 \end{align} $$ And for large enough $n$: $$ \begin{align} \mathbb P(|X-X_n|>\epsilon) = \mathbb P(X>n)= \frac1n\to 0 \end{align} $$ And finally: $$ \mathbb E(X) = \int_1^{+\infty} \frac{dx}x =\lim_{x\to+\infty}\ln x= +\infty $$
Essentially, you choose beforehand $X$ which has infinite expected value. You then cutoff its extreme values and "redirect" them to a negative value tuned so that the expected value is zero. You then let the cutoff go to infinity.
Hope this helps.
Consider a Cauchy r.v. $X$ with density $$f(x)=\frac1\pi\frac1{1+x^2}$$
Define $X_n=X\mathbb{1}_{(0,n)}(X)+X\mathbb{1}_{(-a_n,-n)}(X)$ with $a_n>0$ to be determine so that $E[X_n]=0$:
$$\frac1\pi\Big(\int^{-n}_{-a_n}\frac{x}{1+x^2}\,dx+\int^n_0\frac{x}{1+x^2}\,dx\Big)=\frac1{2\pi}\Big(-\log\big(\frac{1+a^2_n}{1+n^2}\big)+\log(1+n^2)\Big)=0 $$ that is $\log(1+a^2_n)=2\log(1+n^2)$, and so $a_n=\sqrt{n^2+2n}$
Observe that $X_n\xrightarrow{n\rightarrow\infty} X_+$ a.s, and thus in probability.