TITLE: Verifying an inequality under certain conditions.
QUESTION [6 upvotes]: Let $\mu$ be a positive measure on a measure space $\Omega$. How can I show that for all non-negative measurable $f$ $$\exp\left(\int_\Omega f \; d\mu \right) \leq \int_\Omega \exp(f) \; d\mu$$
the above inequality is valid if $\mu(\Omega) = 1$ and not valid if $\mu(\Omega) \neq 1.$ 
This is what I've been able to come up with.  
Since $\exp$ is convex, if $\mu(\Omega) =1$, then the inequality is just Jensen's inequality. Do I have to say more? How do I show that the inequality is not valid if $\mu(\Omega) \neq 1$.

REPLY [3 votes]: What we have to show is that the following assertions are equivalent:
(i) For all $f$ measurable and non-negative, we have $\exp\left(\int_{\Omega} fd\mu\right)\leq \int_{\Omega}\exp(f)d\mu$;
(ii) $\mu(\Omega)=1$. 
The direction $(ii)\Rightarrow (i)$ is as noticed a consequence of Jensen inequality. To see the converse, take $f:= 1$, which is non-negative and mesurable. Let $m:=\mu(\Omega)$. We have $\exp(m)\leq m\cdot e$. Let $h(t)=e^t-te$. Then $h'(t)=e^t-e$, so $h$ is non-negative and $h(t)=0$ if and only if $t=1$, so $m=1$.