Let $f_n \to f$ in $L^1[0,1]$, prove $\int_{0}^{1}e^f\leq \liminf _{n \to \infty}\int_{0}^{1}e^{f_n}$

Question

I am a little clueless how to prove this. My observations: We can use Jensen's inequality here $$e^{\int_{0}^{1}f}\leq \int_{0}^{1}e^f,e^{\int_{0}^{1}f_n}\leq \int_{0}^{1}e^{f_n} $$ However, I do not see his that helpful. I smell fatous but do not know how to apply it here. All I got from it is that

$$e^{\int_{0}^{1}\liminf f}\leq e^f$$

Answer 1

1. By Markov's inequality, we have $f_n \to f$ in measure.

2. Since $t \mapsto e^t$ is a continuous function, it follows that $e^{f_n} \to e^f$ in measure (this is sometimes called the "continuous mapping theorem").

3. Apply Fatou's lemma.