Product of $L^{1}$ Function and Exponentially Integrable Function 
Problem. Let $g\geq 0$ be in $L^{1}[0,1]$ and suppose that $\int gfdx<A$ whenever $\int e^{f}dx\leq 1$.
  
  
*
  
*What can one say about $|\{g>\lambda\}|$ for $\lambda\gg 1$.
  
*Is $g\in L^{2}[0,1]$?
  

The above is an old qual problem with which I've been grappling. There's a hint for the first question, which says to consider $f=c\chi_{E}$. If such an $f$ satisfies $\int e^{f}dx\leq 1$, then
$$\int e^{f}dx=e^{c}|E|+|E^{C}|\leq 1\Leftrightarrow e^{c}\leq 1$$
I.e. $c\leq 0$. I'm not seeing how this helps me, though.
Obviously, by Chebyshev, $|\{g>\lambda\}|\leq\|g\|_{L^{1}}\lambda^{-1}$, but I'm guessing the point of the first question is to determine if we can get better decay than this. For the second question, decay better than $\lambda^{-2}$ would answer in the affirmative; but I don't have a counterexample for otherwise. Any help?

Maybe it is better to use the function $f=a\chi_E-b$. In order to satisfy $\int e^f\leqslant 1$, we should have $b\geqslant \log\left(\left|E\right|\left(e^a-1\right)+1\right)$. Therefore, for any measurable set $E$ and $a\gt 0$, the function $f:=a\chi_E-\log\left(\left|E\right|\left(e^a-1\right)+1\right)$ satisfies $\int e^f\leqslant 1$. By assumption, we get
$$a\int_E g\mathrm d\lambda\leqslant A+\log\left(1+\left(e^a-1\right)\left|E\right|\right)\int_{[0,1]}g\mathrm d\lambda.$$
Taking $a:=\log(1+t)$ for $t\gt 0$, this becomes 
$$\tag{*}\log(1+t)\int_E g\mathrm d\lambda\leqslant A+\log\left(1+t\left|E\right|\right)\int_{[0,1]}g\mathrm d\lambda.$$
Define $\varepsilon(u):=\left|\{g\gt u\}\right|$. We have (taking $t=1/\varepsilon(u)$ and $E:=\{g\gt u\}$)
$$-u\cdot \varepsilon(u)\cdot\log\left(\varepsilon(u)\right)\leqslant K$$
for some $K$ independent of $u$.
Maybe a good optimization in (*) will give better results.
