Question: Suppose that I know that for some $\theta>0$,

$$\mathbb{E}[e^{\theta X}]<\infty$$

How can I show that $\mathbb{E}[X^{+}]<\infty$?

I think I can use the property that $e^{x}$ grows eventually faster than any linear function of $x$, but don't how to formally use this. Please help.

Answer: 
For any $N\geq 0$, let $X\_N$ be $0$ when $X<N$, and $X$ when $X\geq N$. Since $X^+-X\_N$ is bounded for each $N$, it suffices to show that $E(X\_N)<\infty$ for some $N$. But there exists an $N$ such that $e^{\theta t}\geq t$ for all $t\geq N$. For such an $N$, we have $e^{\theta X}\geq X\_N$ everywhere, so $E(X\_N)\leq E(e^{\theta X})<\infty$.
Let $A=\{\omega|X(\omega) \ge 0 \}$. Also note that $e^t \ge 1+t$ for all $t$.

Since $e^{\theta X(\omega)} >0$ for all $\omega$, we have
$E[e^{\theta X}] \ge E[e^{\theta X } 1\_A]\ge E[(1+\theta X) 1\_A] = 1+ \theta E[X 1\_A]$.

Hence $E[X 1\_A] < \infty$.