$\mathbb L^1 +$ a.s. convergence of sequence $(X_n)$ does not imply $\sup(x_n)$ is integrable

Question

I'm looking for a counterexample. The setting is this: Given an probability space $(\Omega,\mathbb{F},\mathbb{P}) $, I look for sequence of random variables $(X_n)_n$ and a random variable $X$, all in $L^{1}(\Omega,\mathbb{F},\mathbb{P})$ s.t.:

$(X_n)_n \rightarrow X$ both a.s. and in $L^{1}$ such that $\sup_{n}X_n \notin L^{1}$.

I need this counterexample to show that it does not hold true that in the above setting with $\mathbb{G} \subset \mathbb{F}$ a subsigma algebra: $\mathbb{E}[X_n|\mathbb{G}] \rightarrow {E}[X|\mathbb{G}]$ a.s.

Thanks a lot.

Answer 1

But if you want to do it your way, let $A_i$ be a disjoint collection of sets with $P(A_i) = c \frac 1 {i^2}$, normalized to sum to something less than 1. Given any positive $\mathbb L^1$ convergent, a.e. convergent sequence $X_n$ let $Y_n = X_n + n1_{A_n}$. Since $\mathbb E( n1_{A_n}) \rightarrow 0$, it still converges in $\mathbb L ^1$, but $sup Y >\sum n 1_{A_n}$ which is not integrable, and it still converges pt-wise, obviously, on the complement of $\cup _n A_i$