Proof about the uniqueness of the Laplace transform in a general setting

Question

Reading the book A Graduate Course on Statistical Inference of Bing Li and Jogesh Babu in page 43 it is asserted that

$$ \left(\forall \theta \in \mathbb{R}^k:\int_{\mathbb{R}^k}g(t)e^{\theta \cdot t}d\nu(t)=0\right)\implies g=0,\quad\nu\text{-a.e.}\tag1 $$

where in the LHS $\nu$ is an arbitrary $\sigma$-finite measure and $g$ is integrable respect to this measure. The book says that the implication is consequence of the uniqueness of the Laplace transform, however I don't found a proof about this uniqueness in the general setting of an arbitrary measure.

Can someone give me a proof, or reference that includes a proof, of the uniqueness of the Laplace transform in this context? Thank you in advance.

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ADDITION: letting $\theta\in \mathbb C ^k$ it can be seen that if $t\mapsto g(t)e^{\theta \cdot t}$ is $\nu $-integrable in some open subset $\Theta $ the function defined by

$$ f(\theta ):=\int_{\mathbb{R}^k}g(t)e^{\theta \cdot t}d\nu(t)\tag2 $$

is analytic in $\Theta $ (i.e. it is complex differentiable). Therefore it follows easily that

$$ \int_{\mathbb{R}^k}g(t)e^{\theta \cdot t}d\nu(t)=0\implies \int_{\mathbb{R}^k}g(t)p(t)e^{\theta \cdot t}d\nu(t)=0\tag3 $$

for any arbitrary polynomial $p$ (of real or complex coefficients). Maybe this can be used to prove (1), however I dont see by now how.

Answer 1

I found a beautiful proof in the book Decomposition of Multivariate Distributions of Roger Cuppens that I will essentially reproduce here for completeness. We start from the point that $\theta \mapsto \int_{\mathbb{R}^k}g(t)e^{\theta \cdot t}d\nu(t)$ is an analytic function in the whole $\mathbb{C}^k$, and so it is zero everywhere because its zero in some open subset. Then for any trigonometric polynomial $T$ we have that $\int_{\mathbb{R}^k}Tg\, d\nu=0$, so

$$ \int_{\mathbb{R}^k}T\,d\mu^+=\int_{\mathbb{R}^k}T\,d\mu^-\tag1 $$

where $d\mu^+:=g^+\mathop{}\!d \nu,\,d\mu^-:=g^-\mathop{}\!d \nu$ where $g^+:=\max\{g,0\},\,g^-:=-\min\{g,0\}$, so $\mu^+$ and $\mu^-$ are non-negative finite measures. Also we will use $|\mu|:=\mu^++\mu^-$, that is, $|\mu|$ is the total variation of $\mu$.

Let $K$ any compact set, then $\mathbf{1}_{K}$ can be approximated by continuous functions with compact support, as the continuous functions are dense in $L^1(|\mu|)$ so, for arbitrarily small $\epsilon >0$ suppose that $\int_{\mathbb{R}^k}|f-\mathbf{1}_{K}|\,d|\mu|<\epsilon$ for some continuous function $f$ with values in $[0,1]$. Now define a periodic version of $f$, namely $f_p=f$ in the support of $f$ such that $f_p$ have period $p$ in every orthogonal direction of $\mathbb{R}^k$. Then clearly $f_p\to f$ pointwise as $p\to \infty $, and due to the Stone-Weierstrass theorem for every $f_p$ there is a trigonometric polynomial $T_p$ such that $\|T_p-f_p\|_{\infty }<p^{-1}$. Therefore if $P\subset \mathbb{R}^k$ is a cube of side lengths $p$ so there is a full period of $T_p$ there and we choose it such that $\operatorname{supp}(f)\subset P$, then $\|\mathbf{1}_{P}T_p-f\|_\infty <p^{-1}$.

Then for $p^{-1}<\epsilon $ we have that $$ -\epsilon \mu^+(\mathbb{R}^k)+\int_{\mathbb{R}^k}f d\mu^+\leqslant \mu^+(K)\leqslant \epsilon \mu^+(\mathbb{R}^k)+\int_{\mathbb{R}^k}f d\mu^+\\ \implies -2\epsilon\, \mu^+(\mathbb{R}^k)+\int_{P}T_p d\mu^+\leqslant \mu^+(K)\leqslant 2\epsilon\, \mu^+(\mathbb{R}^k)+\int_{P}T_p d\mu^+\tag2 $$

As the same holds for $\mu^-$ then

$$ -2\epsilon \,|\mu|(\mathbb{R}^k)+\int_{P}T_p\,d\mu\leqslant \mu(K)\leqslant 2\epsilon \,|\mu|(\mathbb{R}^k)+\int_{P}T_p\,d\mu\\ \iff -2\epsilon \,|\mu|(\mathbb{R}^k)-\int_{P^\complement }T_p\,d\mu\leqslant \mu(K)\leqslant 2\epsilon \,|\mu|(\mathbb{R}^k)-\int_{P^\complement }T_p\,d\mu\\ \implies |\mu(K)|\leqslant 2\epsilon |\mu|(\mathbb{R}^k)+\left| \int_{P^\complement }T_p\,d \mu \right|\tag3 $$

Finally, as $|\mu|(\mathbb{R}^k)<\infty $ then for every $\epsilon >0$ there is some $N>0$ such that $|\mu|(P^\complement )<\epsilon $ when $p\geqslant N$, then for large enough $p$ we have the bound $\left|\int_{P^\complement }T_p\,d \mu \right|\leqslant \epsilon (\epsilon +1)$. This together with (3), and as $\epsilon >0$ is arbitrary, we finally find that $\mu(K)=0$ for every compact set $K\subset \mathbb{R}^k$. As $\mu$ is a finite Borel measure its regular so every Borel set $\mathbb{R}^k$ can be approximated from below for compact sets, therefore $\mu$ is the zero measure.∎

Answer 2

The cleanest proof of uniqueness uses the transform $u = \exp(-t),$ then the uniqueness for Laplace transform reduces to proving that if the integral of $p(x) h(x)$ on $[0, 1]$ for any polynomial $p(x),$ then $h(x) = 0.$ That can be proved in many ways (Stone-Weierstrass, Korovkin...).