To prove that $X=\infty$ a.s. given that $X$, $Y$ are independent and $X \sim X + Y$. I saw this statement when reading a textbook and I think it's correct intuitively. However I am not able to give a rigorous proof of the statement. To be more specific, what I want to prove is the following:

Let $X$ and $Y$ be two independent random variable, such that $X \in [0, \infty]$ and $Y$ has the standard normal distribution. If $X$ and $X + Y$ has the same distribution, then $X = \infty$ almost surely.

 A: Since $X \stackrel{d}{=} X+Y$ and $Y$ takes real values, it follows that
$$X 1_{\{|X|<\infty\}} \stackrel{d}{=} (X+Y) 1_{\{|X+Y|<\infty\}} = (X+Y) 1_{\{|X|<\infty\}}. \tag{1}$$
By the independence of $X$ and $Y$, we have
$$\begin{align*} \mathbb{E}\exp(i \xi (X+Y) 1_{\{|X|<\infty\}}) &= \mathbb{E} \left( 1_{\{|X|=\infty\}} + 1_{\{|X|<\infty\}} \exp(i \xi (X+Y)) \right) \\ &= \mathbb{P}(|X|=\infty) + \mathbb{E}(e^{i \xi X} 1_{\{|X|<\infty\}}) \mathbb{E}\exp(i \xi Y)\\ &= \mathbb{P}(|X|=\infty) + \mathbb{E}(e^{i \xi X} 1_{\{|X|<\infty\}}) \exp(-\xi^2/2)\end{align*}$$
On the other hand, $(1)$ gives
$$\begin{align*} \mathbb{E}\exp(i \xi (X+Y) 1_{\{|X|<\infty\}}) &\stackrel{(1)}{=} \mathbb{E}\exp(i \xi X 1_{\{|X|<\infty\}}) \\ &= \mathbb{P}(|X|=\infty) + \mathbb{E}(e^{i \xi X} 1_{\{|X|<\infty\}}) \end{align*}$$
Thus,
$$\mathbb{P}(|X|=\infty) + \mathbb{E}(e^{i \xi X} 1_{\{|X|<\infty\}}) =  \mathbb{P}(|X|=\infty) + \mathbb{E}(e^{i \xi X} 1_{\{|X|<\infty\}}) \exp(-\xi^2/2)$$
i.e.
$$\mathbb{E}(e^{i \xi X} 1_{\{|X|<\infty\}}) =  \mathbb{E}(e^{i \xi X} 1_{\{|X|<\infty\}}) \exp(-\xi^2/2)$$
This identity can only hold true if
$$\mathbb{E}(e^{i \xi X} 1_{\{|X|<\infty\}})=0$$
for all $\xi \in \mathbb{R} \backslash \{0\}$. Letting $\xi \to 0$ we conclude
$$\mathbb{P}(|X|<\infty)=0.$$
Remark: Note we did not use the assumption that $X$ is non-negative; the above reasoning works for any random variable $X$ which satisfies $X  \stackrel{d}{=} X+Y$ for some independent $Y \sim N(0,1)$.
A: For $s > 0$, we know that $\mathbb{E}\big[e^{-sX}\big]$ lies in $[0, 1]$ and satisfies
$$ \mathbb{E}\big[e^{-sX}\big]
= \mathbb{E}\big[e^{-s(X+Y)}\big]
= e^{s^2/2}\mathbb{E}\big[e^{-sX}\big]. $$
So we have $\mathbb{E}[e^{-sX}] = 0$. Now
$$ \mathbb{P}[X < \infty] = \mathbb{E}\big[\mathbf{1}_{\{X < \infty\}}\big] \stackrel{\text{(Fatou)}}{\leq} \lim_{s\downarrow 0} \mathbb{E}\big[e^{-sX}\big] = 0. $$
A: Note, $X=\infty \Leftrightarrow X + Y = \infty \, a.s.$ hence $Z=X|X<\infty$ has the same distribution as $Z+Y$, checking the characteristic functions means that the characteristic function of $Z$ is zero everywhere except $s=0$ which correspond to a uniform meassure, $\mu$ on $R$. but for all sets $A$ on R we must have $\mu(A)=0$ so this can only be the zero meassure. Hence all mass of $X$ is at $\infty$.
