Let $X$ be a random variable, and $L_X: \mathbb{R} \rightarrow > [0,\infty]$ defined as $$ L_X(\lambda) = \int_{\mathbb{R}}e^{- \lambda x}P_x(dx)$$ The random variable $X$ is chosen such that $L_X(\lambda) < \infty$ for $\lambda$ in a neighbourhood of $0$.
I have to show that if we put $A = \{ \lambda \in \mathbb{R}, L_X(\lambda) < \infty \}$, then $\mathring{A}$ is an open interval of $\mathbb{R}$ and then I have to work with elements belonging to $\mathring{A}$, thus I suppose being it an empty set isn't a possibility. This rather perplexes me, as because we integrate on $\mathbb{R}$, I believe that $L_X(\lambda)$ will explode for any $\lambda \in \mathbb{R}$, whereas if we would integrate on only $\mathbb{R}^+$, we wouldn't have such problem. So I wonder, does the definition of $L_X$ has any sense in our context?
Suppose $\lambda_1 \in A$, $\lambda_2 \in A$ and $\lambda_1 < \mu <\lambda_2$ Since $e^{-\mu x} \leq e^{-\lambda_1 x}$ if $x \geq 0$ and $e^{-\mu x} \leq e^{-\lambda_2 x}$ if $x < 0$ we get $e^{-\mu x} \leq e^{-\lambda_1 x} +e^{-\lambda_1 x}$ always. Integrating we find that $\mu \in A$. This property means $A$ is an interval so its interior is an open interval. (In the comments of the OP the assumption that $A$ contains an interval around $0$ has been ignored so those comments are irrelevent).