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Let $\mu$ and $\nu$ be nonnegative Borel measures on $[0,+\infty)$ and let measure $\mu$ be finite, i.e. $\mu \{ [0,+\infty) \} < \infty$. For any $p > 0$ the equality holds $$ \int\limits_{0}^{\infty} e^{-px} \mu(dx) = \int\limits_{0}^{\infty} e^{-px} \nu(dx). $$ Passing to the limit $p \to +0$ and using Lebesgue dominated convergence theorem we receive $$ \int\limits_{0}^{\infty} \mu(dx) = \lim\limits_{p \to +0} \int\limits_{0}^{\infty} e^{-px} \nu(dx) < \infty. $$ Is it true that measure $\nu$ is finite in general? If I'm not mistaken we can use monotone convergence theorem to pass to the limit under the integral sign. More precisely, $e^{-p_n x}$ is a monotone nondecreasing sequence of functions if $p_n$ is a decreasing sequence with zero limit. Furthermore, $e^{-p_n x} \to 1$. Then $1$ is $\nu$-integrable and $$ \int\limits_{0}^{\infty} \mu(dx) = \lim\limits_{p \to +0} \int\limits_{0}^{\infty} e^{-px} \nu(dx) = \int\limits_{0}^{\infty} \lim\limits_{p \to +0} e^{-px} \nu(dx) = \int\limits_{0}^{\infty} \nu(dx) $$

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up vote 3 down vote accepted

Yes: by Fatou's lemma, $$\nu([0,+\infty))=\int_{[0,+\infty)}1d\nu(x)=\int_{[0,+\infty)}\lim_{n\to +\infty}e^{-n^{-1}x}d\nu(x)\leqslant \liminf_{n\to +\infty}\int_{[0,+\infty)}e^{-n^{-1}x}d\nu(x),$$ which gives $$\nu([0,+\infty))\leqslant \liminf_{n\to +\infty}\int_{[0,+\infty)}e^{-n^{-1}x}d\mu(x)\leqslant \mu([0,+\infty)).$$ And actually, the masses are equal.

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