Uniform integrability and $L^1$ convergence of $(1/X_n)$

Question

Let $X_n \to c > 0$ almost surely, where $c$ is a constant and $X_n > 0$ for all $n$. Also, let $(X_n)$ be uniformly integrable, so in particular $X_n \to c$ in $L^1$.

Question:

Do we have uniform integrability of $(1/X_n)$ and thus convergence in $L^1$ of $1/X_n$ to $1/c$?

Thanks.

Answer 1

Answer is no: On $[0,1]$ define $X_n(t)=1 $ if $\frac{1}{n}\le t\le 1$ and $X_n(t)=\frac{1}{n^2}$ if $t\lt \frac{1}{n}$. This sequence satisfies the given conditions (with $c=1$) but $||1/X_n||_{L_1} \gt n$ for all $n$ so it's not uniformly integrable and does not converge on $L_1$.