Does there exist $\{X_n\}$ for which $\liminf_{n\to \infty}X_n$ doesn't exist but the negative parts $\{X^-_n\}$ are uniformly integrable?

Question

Are there random variables $\{X_n\}_{n\ge 1}$ for which the expected value of $ \liminf\limits_{n\to \infty} X_n $ doesn't exist but the negative parts $\{X^-_n\}_{n\ge 1}$ are uniformly integrable?

How can we prove analytically that there exists?

Answer 1

If I understand the question correctly, the classic example of a sequence of functions (random variables) that converges in measure to zero, but does not converge pointwise to zero (for any $x$), would be an example of what you asked.

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Let $X_n=-\sqrt{\frac{1}{\mu(I_n)}}\cdot\chi_{I_n}(x)$, where $$\chi_{I_n}(x)= \begin{cases} 1 &x \in I_n \\ 0 & x \notin I_n \end{cases}$$

And $\{I_n\}_{n\geq1}$ is $[0,\frac{1}{2}],[\frac{1}{2},1],[0,\frac{1}{3}],[\frac{1}{3},\frac{2}{3}],[\frac{2}{3},1],[0,\frac{1}{4}],[\frac{1}{4},\frac{2}{4}]...$

You get that $\liminf\limits_{n\to \infty} X_n = -\infty$ and therefore its expectation does not exist.

$$\lim\limits_{n\to\infty}\int\limits_0^1 X_nd\mu = 0$$

so you can prove they are uniformly integrable.