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Suppose that $\mu$ is a finite Borel measure on $[0,∞)$. Prove that $\int_{(0, \infty]} e^{\alpha x} d\mu < \infty$ for some $\alpha > 0$ if and only if there exist $c, C > 0$ such that $\mu([t, ∞)) \le Ce^{−ct}$ for all $t > 0$.

$\Rightarrow$: This is the easier direction. Fix $C > 0$. Suppose for contradiction that for every $c > 0$ there exists a $t > 0$ such that $\mu([t, ∞)) > Ce^{−ct}$. Then for any $\alpha$, $\int_{(0, \infty]} e^{\alpha x} d\mu \ge \int_{(t, \infty]} e^{\alpha x} d\mu \ge \text{min}_{(t, \infty]} e^{\alpha x} \mu (t, \infty] > e^{\alpha t} Ce^{−ct} = C e^{(\alpha - c)t}$. If $\alpha \ge c$ then $\int_{(0, \infty]} e^{\alpha x} d\mu > C$ for arbitrary $C$, and we derive a contradiction. I am not sure how to handle to case when $\alpha < c$.

$\Leftarrow$: This direction I only have the general idea. Choose $\alpha < c$. $\int_{(0, \infty]} e^{\alpha x} d\mu = \int_{(0, t]} e^{\alpha x} d\mu + \int_{(t, \infty]} e^{\alpha x} d\mu $. If I can show that the radon-nikodym theorem applies in this case, then $= \int_{(0, t]} e^{\alpha x} d\mu + \int_{(t, \infty]} e^{\alpha x} \frac{d\mu}{dm} d\mu$ where m is the Lebesgue measure and $\frac{d\mu}{dm}$ is the radon-nikodym derivative such that $\mu(E) = \int_E \frac{d\mu}{dm} dm$. I know that $\frac{d\mu}{dm}$ is the derivative of the distribution function of $\mu$. If I can then show that $\frac{d\mu}{dm}(x) \le Ce^{-c x} $, I will have the bound.

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1 Answers1

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WLOG let $\mu$ be a probability measure.

Let $$\int_0^\infty e^{\alpha x}\mu(dx)<\infty$$ for some $\alpha>0$. Then by Markov's inequality (well extended version with $\varphi(x)=e^{\alpha x}$) we get for $t\geq 0$ that $$\mu([t,\infty))\leq \frac{1}{e^{\alpha t}}\int_0^\infty e^{\alpha x}\mu(dx)=Ce^{-\alpha t},$$

where $C=\int_0^\infty e^{\alpha x}\mu(dx)<\infty$.

Conversely, let $\mu([t,\infty))\leq Ce^{-ct}$ for some $c,C>0$. Then note that by tail integral for expectation we have

$$\int_0^\infty e^{\alpha x} \mu(dx)=\int_0^\infty \mu([e^{\alpha x},\infty))dx.$$

Therefore $$\int_0^\infty e^{\alpha x} \mu(dx)\leq \int_0^\infty C e^{-c e^{\alpha x}}dx<\infty.$$