I'm working on a paper, and I don't know if there is some kind of typo or if I just don't get what seems obvious to the author.
Note : I'll be working with probabilities, but I guess this would be the same in a finite measure framework.
So we denote by W the Wiener space, $\mu$ the Wiener measure, and let $f$ be a random variable (ie : measurable function on W). We denote $f^+$ and $f^-$ the positive and the negative parts.
Assume that there exist $c>1$ such that $\mathbb{E}[\exp(cf^-)]<\infty$ and $\varepsilon>0$ such that $f\in L^{1+\varepsilon}(\mu)$. Then : $$\mathbb{E}[(|f|+1)e^{-f}]<\infty$$
Since $e^{-f}\leq e^{f^-}$, we only need to consider $\mathbb{E}[|f|e^{f^-}]$. If $c$ and $1+\varepsilon$ are conjuguate in the sense of the Hölder inequality then fine, but otherwise, is there any obvious reason why this would hold true ?
Second point : same question about $\mathbb{E}[(|f|+1)e^{-f}]$ but assuming that $\mathbb{E}[f^- e^{f^-}]<\infty$, $f\in L^{1+\varepsilon}(\mu)$ and $\mathbb{E}[e^{-f}]<\infty$
I think the argument should start like this: let $$ M = \sup_{x>0} (1+x) e^{-(c-1) x} .$$ Clearly $M < \infty$. Then $$(1+f^-) \exp(f^-) \le M \exp(c f^-) .$$ Therefore if $E[\exp(c f^-)] < \infty$, then $E[(1+f^-) \exp(f^-)] < \infty$.
I think the rest of the argument is similar. But I must confess that I don't see why $f \in L^1$ isn't enough.