Sufficient conditions for L1 convergence of exponentials

Question

Suppose $(X,B,m)$ is a finite measure space and $(f_n)$ is a sequence of functions converging almost surely and in $L^2(X,m)$. Moreover, I know that for every $n$, $\int e^{f_n(x)} m(dx)<\infty$. What would be a natural condition to ensure that functions $$ F_n(x)=\frac {e^{f_n(x)} }{\int e^{f_n(x)} m(dx) } $$ converge in $L^1(X,m)$? Convergence of $f_n$'s in $L^p$ for all $p\ge 1$, seems sufficient.

Answer 1

Once we have convergence $F_n \to F$ in measure (this is the case in your situation), the natural condition for convergence in $L^1$ is uniform integrability. This is equivalent to convergence $||F_n||_1 \to ||F||_1$.

Assuming that the sequence $(f_n)$ converges in every $L^p$ does not ensure that the sequence $(F_n)$ converges in every $L^1$.

For example, assume that $m$ is a probability measure, that each $f$ is an exponential random variable with parameter $1$ and that $f_n = (1+n^{-1})f$ for every $n \ge 1$. Then the sequence $(f_n)$ converges to $f$ in every $L^p$. Yet $$E(e^{f_n}) = \int_0^\infty e^{(1+n^{-1})x}e^{-x}dx = n \to +\infty,$$ so $F_n \to 0$ as $n \to +\infty$, although $E(F_n)=1$ for every $n$.