Let $A$ be a random variable and $P$ be a probability measure. For some real $\delta>0$, is there a simple proof that $E[\exp(|A|)] \leq \exp(\delta) + \exp(\delta)P(|A|>\delta) + \int_\delta^\infty\exp(x)P(|A|>x)dx$?
This statement was written as obvious by Capitaine, page 196 and Ikeda and Watanabe page 450, but I've failed to get any progress in understanding it. Capitaine says it follows from Fubini's theorem. I think it is not necessary to assume that $A$ admits a density.
By definition,
$$\begin{align*} \mathbb{E}e^{|A|} &= \int 1_{\{|A| \leq \delta\}} \underbrace{e^{|A|}}_{\leq e^{\delta}} \, d\mathbb{P} + \int 1_{\{|A|> \delta\}} \cdot e^{|A|} \, d\mathbb{P} \\ &\leq \underbrace{e^{\delta} \cdot \mathbb{P}(|A| \leq \delta) + e^{\delta} \cdot \mathbb{P}(|A|>\delta)}_{=e^{\delta} \leq e^{\delta}+e^{\delta} \cdot \mathbb{P}(|A|>\delta)} + \int 1_{\{|A|> \delta\}} \cdot (e^{|A|}-e^{\delta}) \, d\mathbb{P} \tag{1} \end{align*}$$
Now note that
$$\begin{align*} \int_{\delta}^{\infty} e^x \cdot \mathbb{P}(|A|>x) \, dx &= \mathbb{E} \left( \int_{\delta}^{\infty} e^x \cdot 1_{\{|A|>x\}} \, dx \right) \tag{2} \end{align*}$$
by Fubini's theorem where
$$\begin{align*} \int_{\delta}^{\infty} e^x \cdot 1_{\{|A|>x\}} \, dx &= \begin{cases} \int_{\delta}^{|A|} e^x \, dx = e^{|A|}-e^{\delta} & |A|>\delta \\ 0 & |A| \leq \delta \end{cases} \\ &= (e^{|A|}-e^{\delta}) \cdot 1_{\{|A|>\delta\}} \tag{3} \end{align*}$$
Therefore, the claim follows by combining (1)-(3).