I'm somewhat stuck on understanding, what seems to be a kind of elementary estimate. Let
- $X\geq 0$ with $\mathbb{E}(X)=1$.
- $f:\mathbb{R}^+ \to \mathbb{R} $ and $f(x)\geq-c>-1$.
- Let $A$ be some arbitrary event from the probability space.
How can I show that
$$\mathbb{E}\left[\left(f(X)\right)^+\mathbf1_A\right]\leq c+\mathbb{E}\left[f(X)\right]$$
EDIT: $(\cdot)^+$ is defined as $\max{(\cdot,0)}$.
Write $f(X) = f(X)^+ - f(X)^-$. Then, observe that the bound $f \geq -c$ implies $f(X)^- \leq c$ a.s., so that $$ \mathbb{E}[f(X)] = \mathbb{E}[f(X)^+ - f(X)^-] = \mathbb{E}[f(X)^+] - \mathbb{E}[f(X)^-] \geq \mathbb{E}[f(X)^+] - c $$ and reorganizing the terms give what you want: $\mathbb{E}[f(X)^+] \leq \mathbb{E}[f(X)] + c$.
Note that this does not require the assumption $X \geq 0$ (except for $f(X)$ to be defined, if you really want the domain of $f$ to be $\mathbb{R}_+$) nor $\mathbb{E} X =1$.
Now, you can generalize to your specific expression by observing that $f(X)^+\mathbf{1}_A \leq f(X)^+$, since $f(X)^+ \geq 0$ a.s.: so that $$\mathbb{E}[f(X)^+\mathbf{1}_A] \leq \mathbb{E}[f(X)^+] \leq \mathbb{E}[f(X)] + c.$$