# Levy process:2:$E\left[\exp\left(\theta^*X_\tau -\Psi(\theta^*)\tau\right)1_{\{\tau < \infty\}}\right] = e^{\theta^*x}P\left( \tau < \infty \right)$

This a sequel of this exercise, which label is repeated below.
My question follows.

Exercise
Let $X_t$ be the compound Poisson process $$X_t = t - \sum_{i=1}^{N_t} \xi_i, \tag{1}$$ where $N$ is a Poisson process with rate $\lambda$ and the $\xi_i$ are i.i.d., positive, with common distribution $F$.
The characteristic exponent of $X_t$ is $$\Psi(\theta) = \theta - \lambda \int_{(0, \infty)} \left( 1 - e^{-\theta x}\right)F(dx). \tag{2}$$ Assume $\Psi$ has a root $\theta^* \ne 0$. Define the stopping time $$\tau = \inf_{t > 0} \left\{ X_t > x \right\} , x > 0. \tag{3}$$
Show $$E\left[\exp\left(\theta^*X_\tau -\Psi(\theta^*)\tau\right)1_{\{\tau < \infty\}}\right] = e^{\theta^*x}P\left( \tau < \infty \right). \tag{4}$$

This is exercise 1.9 in Kyprianou, Fluctuation of Levy Process.
The solution is given in the book.

Question
One of the step to get (4), is to use the fact that $$M_t = e^{\theta^*X_{t \land \tau} -\Psi(\theta^*)(t \land \tau)}, \tag{A}$$ is a martingale, to get $$E\left( M_t \right) = M_0 = 1. \tag{B}$$ As $\Psi(\theta^*)=0$, taking the limit as $t$ goes to $\infty$ in (B) gives $$E\left[ e^{\theta^*X_\tau}\right]= 1. \tag{C}$$ Now, infer that $$E\left[e^{\theta^*X_\tau }1_{\{\tau < \infty\}}\right] = 1 . \tag{D}$$

I don't see how to go from (C) to (D).
I know $$E\left[ e^{\theta^*X_\tau}\right]=E\left[e^{\theta^*X_\tau }1_{\{\tau < \infty\}}\right] + E\left[e^{\theta^*X_\tau }1_{\{\tau = \infty\}}\right], \tag{E}$$ but why would the second term be zero?

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