laplace transform probability distribution not concentrated on 0

Question

This seems intuitively obvious but how to prove that $\hat{\mu} < 1,$ when $\theta >0$ and $\mu$ is a probability measure not concentrated at $0,$ where $\hat{\mu}$ is defined as below $$\hat{\mu}(\theta) = \int\limits_{0}^{\infty}e^{-\theta x}d\mu(x)$$

Answer 1

If $\mu(\{0\})<1$, then as $e^{-\theta x}<1$ for $\theta,x>0$ we have \begin{align} \hat\mu(\theta) &= \mu(\{0\}) + \int_{(0,\infty)} e^{-\theta x}\ \mathsf d\mu(x)\\ &< \mu(\{0\}) + \int_{(0,\infty)}\ \mathsf d\mu(x)\\ &=\mu(\{0\}) + \mu((0,\infty))\\ &=\mu([0,\infty))\\ &\leqslant 1. \end{align}