# how to prove that $\mu$ is borel-finite

I have a Borel non-decreasing measure $\mu$ such that

$$\int_{-\infty}^{+\infty}\left|\sum_{i=1}^n \xi_i e^{-y_i t}\right|^2 d\mu(t)\geq 0$$ and finite for every $n\in\mathbb{N}$, every $\{\xi_i\}_{i=1\ldots n}$ complex sequence and every $\{y_i\}_{i=1\ldots n}$ real sequence different from (0,...0).

I can conclude that $\mu(\mathbb{R})>0$, but can I say that $\mu(\mathbb{R})<+\infty$? I would like to have finite $\mu$ in order to prove that $\mu$ is Borel-finite and non-negative.

• What about $n=1$, $\xi_1 = 1$, $y_1 = 0$, then by assumption $\int_{\mathbb R} 1\, d\mu = \mu(\mathbb R)$ is positive and finite?! – martini Jun 18 '13 at 15:52
• And if my assumption is true for every $\{y_i\}_{i=1\ldots n}$ different from the (0,...,0) vector? can you still prove that $\mu$ is finite? – alemou Jun 18 '13 at 16:03
• Let $n = 1$, $\xi_1 = 1$, $y_1 = \pm 1$. Then $|\xi_1 e^{-ty_1}|^2 = e^{\pm 2t}$. As $1 \le e^{2t} + e^{-2t}$, we have $\int_{\mathbb R} 1 \le \int e^{2t} + e^{-2t} < \infty$. – martini Jun 18 '13 at 21:20