Prove that if f(x) is integrable, then so is e^(f(x)).

Question

So here is my question:

I'm working on a homework problem that deals with Jensen's Inequality. It is a rather simple application, I believe, but I'm a little stuck. Here is the problem, along with my work.

Let $f$ be integrable over $[0,1]$. Show that

$\exp\left[\int_0^1 f(x) dx \right] \leq \int_0^1\exp[f(x)]dx$

On the surface, this appeared like a simple application of Jensen's Inequality, since $\phi (x) = e^x$ is convex on $\mathbb{R}$. We know that $f$ is integrable, but it remains to show that $\exp[f(x)]$ is integrable. I know that the composition ($\exp[f(x)]$) is measureable, but I don't feel like that is enough to conclude that it is integrable.

Answer 1

There is no way you can prove $e^{f(x)}$ is integrable just because $f(x)$ is. But it is OK, because if $e^{f(x)}$ is not integrable, then the right hand side is infinite, and the statement is obviously true.

To find a counterexample, look for functions of the form $$ f(x) = 2^{\alpha k} \text{ if $x \in (2^{-k+1},2^{-k}]$} .$$ If $0 \le \alpha < 1$, then this is easily seen to be integrable.

Answer 2

$\exp(f(x))$ is a nonnegative measurable function, so the worst thing that could happen is that the integral is $+\infty$. And this is indeed possible: Let $f(x)=n$ if $\frac 1{n}-\frac1{n^3}<x<\frac1n$ with $n>1$ (and $0$ otherwise). Then $$\int_0^1 f(x)\,\mathrm dx = \sum_{n=2}^\infty \frac1{n^2}=\frac{\pi^2}6-1$$ whereas $$ \int_0^1 \exp{f(x)}\,\mathrm dx = \sum_{n=2}^\infty \frac{e^n}{n^3}$$ is infinite ($e^n>n^3$ for $n$ suficiently big).