If $f$ is Lebesgue integrable on $[0,1]$ then $e^f$ is

Question

Let $f:[0,1] \to \mathbb{R}$ be Lebesgue integrable on $[0,1]$. I want to show that $e^f$ is also integrable.

I see that $e^f$ is nonnegative, so $\int_0 ^1 e^f$ is defined as an exteded real number.

If $f$ were bounded, say $|f| <M$, I would observe that $\int_0 ^1 e^f \leq \int_0 ^1 e^M=e^M< \infty$ and be done. However, $f$ might not be bounded in general.

Answer 1

This is false. Consider $f(x) : = \ln(1/x)$. Then $$ \int_{0}^1 \ln(1/x) \, dx = x + x \log(1/x) \Bigg|_{x = 0}^{x = 1} = 1. $$ Hence $f$ is Lebesgue integrable. On the other hand, $$ \int_{0}^1 e^{f(x)} \, dx = \int_{0}^1 \frac{1}{x} \, dx $$ does not converge.