I am working through Resnick as a self-study, and I have a question about Resnick Probability Path question 5.20. The question is as follows:
For a random variable $X: (\Omega,\mathcal{F},P) \to (\mathbb{R},\mathcal{B}(\mathbb{R})$ with law $F = P \circ X^{-1}$, define the moment generating function $\phi(\lambda)$ by $\phi(\lambda) = E(e^{\lambda X})$. Part (a) asks to prove that $\phi(\lambda) = \int_{R} e^{\lambda x} F(dx)$.
For part a, if $R$ is meant to be $\mathbb{R} = (-\infty,\infty)$, then I believe this is just a simple application of the change of variables. In particular, if we let $h: (\mathbb{R},\mathcal{B}(\mathbb{R})) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ by $h(x) = e^{\lambda x}$, then $h$ is non-negative, and measurable (since it is continuous). So a direct application of the change of variables gives $\int_{\Omega} h \cdot X dP = \int_{\mathbb{R}} h(x) dF(x) = \int_{\mathbb{R}} e^{\lambda x} dF(x)$
So my first question is this: is $R$ meant to be $\mathbb{R}$ (i.e the real line)? Or does $R$ refer to something else?
For part b, $\Lambda = \{\lambda \in R: \phi(\lambda) < \infty \}$ is defined. And then it is asked that we prove $\phi(\lambda) > 0$ for every $\lambda$ in the interior of $\Lambda$ and that $\phi$ is continuous on the interior of $\Lambda$.
For this part, if $R = (-\infty,\infty)$, then I am confused about why we are asked to prove $\phi(\lambda) > 0$ on only the interior of $\Lambda$. Isn't this true for every $\lambda \in \mathbb{R}$?
Specifically, $\phi(0) = 1$, and for $\lambda > 0$, by properties of the cdf, there must exist $x \in \mathbb{R}$ such that $P(X > x) > 0$. So the function $ e^{\lambda x} \mathbb{I}((x,\infty))$ is a simple function minorizing $e^{\lambda x}$. And $\int_{\mathbb{R}} e^{\lambda x} \mathbb{I}((x,\infty)) dF(x) = P(X > x) > 0$. Thus $\phi(\lambda) > 0$. And if $\lambda < 0$, again by properties of the cdf, there has to exist a $x \in \mathbb{R}$ such that $P(X \leq x) > 0$. So by an analogous argument to the $\lambda > 0$ case, $\phi(\lambda) > 0$ for $\lambda < 0$
So for all $\lambda \in \mathbb{R}$, $\phi(\lambda) > 0$. Is this correct? Or am I missing something?
And then for the other part of (b), it is asked to show $\phi$ is continuous on the interior of $\Lambda$. My argument is as follows:
Let $\lambda$ be in the interior of $\Lambda$. So for some $\delta > 0$, $N_{\delta} (\lambda) = (\lambda - \delta, \lambda + \delta) \in \Lambda$. Now let $x_n \to \lambda$. Without loss of generality we can say $x_n \in N_{\delta}(\lambda)$. For $\lambda > 0$, further assume (wlog) that $\lambda - \delta > 0$. Then for $x \geq 0$, $e^{x_n x} \leq e^{(\lambda + \delta/2)x}$ and for $x < 0$, $e^{x_n x} \leq e^{(\lambda - \delta/2)x}$. So we have that $e^{x_n x} \leq \mathbb{I}(x \geq 0) e^{(\lambda + \delta/2)x}+ \mathbb{I}(x < 0) e^{(\lambda - \delta/2)x}$. Thus
$|e^{x_n x} - e^{\lambda x}| \leq \mathbb{I}(x \geq 0) e^{(\lambda + \delta/2)x}+ \mathbb{I}(x < 0) e^{(\lambda - \delta/2)x}+e^{\lambda x}$
and the right hand side above is integrable, since $\lambda - \delta/2, \lambda+\delta/2, \lambda$ are in $\Lambda$.
Thus since $|e^{x_n x} - e^{\lambda x}| \to 0$ (due to the continuity of e), then applying dominated convergence theorem, $\int_{\mathbb{R}} |e^{x_n x} - e^{\lambda x}| dF(x) \to \int_{\mathbb{R}} 0 dF(x) = 0$. So we conclude that
$|\phi(x_n) - \phi(\lambda)| = |\int_{\mathbb{R}} e^{x_n x} dF(x) - \int_{\mathbb{R}} e^{\lambda x} dF(x)| \leq \int_{\mathbb{R}} | e^{x_n x} - e^{\lambda x} | d F(x) \to 0$
And hence $\phi$ is continuous for $\lambda > 0$ such that $\lambda \in \Lambda$. The case where $\lambda < 0,\lambda \in \Lambda$ is done similarly, and for $\lambda = 0$, also can be done similarly.