I encountered a problem in measure theoretic probability which is as follows:
Theorem
If $\mu_1$ and $\mu_2$ are probability measures on $\mathbb R$ equipped with the Borel $\sigma$-algebra with moment generating functions $\psi_1,\psi_2$ respectively and if $\psi_1(t)=\psi_2(t)<\infty$ for all $E\subset \mathbb R$ and if there exists an accumulation point $x$ of $E$ satisfying $\inf(E)<x<\sup(E)$,then show that $\mu_1=\mu_2$.
I did this problem in the following manner.
My solution
Recall that for a probability measure $\mu$ on $\mathbb R$ the corresponding Moment Generating Function is given by,
$\psi(t)=\int_\mathbb R e^{tx}\mu(dx)$ where $t\in \mathbb R$.Now we note that if $Re(z)\in E$ then $\int_\mathbb R |e^{zx}|\mu(dx)<\infty$(Since $\int_\mathbb R |e^{zx}|\mu(dx)=\int_\mathbb R e^{(Re z)x}\mu(dx)<\infty$ provided $\psi(t)<\infty$ for all $E\subset \mathbb R$ and using the fact that $\mu(\mathbb R)=1$).So the expression $\int_{\mathbb R} e^{zx}\mu(dx)$ is a real number for all $z\in \mathbb C$ such that. $Re(z)\in E$.We define $f(z)=\int_\mathbb R e^{zx} \mu(dx)$ for $z\in E\times \mathbb R\subset \mathbb C$(Vertical strip supported by $E$).Now,$f(z)$ is continuous function of $z$ and $\int_T f(z)dz=\int_T(\int_\mathbb Re^{zx}\mu(dx))dz=\int_\mathbb R(\int_Te^{zx}dz)\mu(dx)=0$ where the interchange of integrals is justified by Fubini's theorem and here $T$ is any triangle in the domain $\Omega=E\times \mathbb R$ and hence by Morera's theorem,we have $f(z)$ is a holomorphic function in $z$.
Now let $\tilde \psi_1(z)=\int_\mathbb R e^{zx}\mu_1(dx)$ and $\tilde\psi_2(z)\int_\mathbb R e^{zx}\mu_2(dx)$ which are both holomorphic functions on $\Omega$ and they agree on the set $E\subset \Omega$ because $\psi_1(t)=\psi_2(t)$ for all $t\in E$ and $E$ has a limit point in $\Omega$ since $E$ is an interval in $\mathbb R$ and $x\in (\inf(E),\sup(E))$ and hence $x\in \Omega$.So,by identity theorem of holomorphic functions $\psi_1=\psi_2$ on whole of $\Omega$ and in particular $\psi_1(it)=\psi_2(it)$ and hence $\phi_1(t)=\phi_2(t)$ for all $t\in \mathbb R$ and hence $\mu_1=\mu_2$ because they have the same CHF.
Is there any mistake in this proof,if there is any,kindly point out.