Let $Z$ be a $[0, \infty)$-valued random variable satisfying $\lambda E[g(Z+1)] = E[Zg(Z)]$ for all indicator functions $g$ of Borel subsets of $[0, \infty)$.
Prove that $\mathcal{L}(Z) = Poisson(\lambda)$.
Hint: Consider $1_{(n,n+1)}$ for $n \in \mathbb{N}_0$.
Proof:
Let $n \in \mathbb{N}_0$ and g = 1_{(n,n+1)}, then holds: \begin{align*} \lambda E[g(Z+1)] = \lambda P[Z+1 \in (n,n+1)] = \lambda (P[Z+1 = n+1] - P[Z+1 = n]) = \lambda P[Z = n] - \lambda P[Z = n-1]). \end{align*} But what is \begin{align*} E[Zg(Z)] = E[Z \cdot 1_{(n,n+1)}(Z)] = ? \end{align*}
Have I got everything right so far? If so, how should I proceed now?
Thanks for the help!
The assertion $$P[Z+1\in(n,n+1)]=P[Z+1=n+1]-P[Z+1=n]\tag1$$ isn't quite correct. To assert the RHS of (1) you would need to prove that $Z$ takes only integer values, and also you would need $(n,n+1]$ on the LHS instead of $(n,n+1)$.
Hint: You should first show that $Z$ takes only integer values. Consider the case $n=0$ first, so try $g:=I_{(0,1)}$ in the identity $\lambda E[g(Z+1)]=E[Zg(Z)]$. You'll get $$\lambda P(0<Z+1<1) = E[ZI_{(0,1)}(Z)].$$ The LHS equals zero because the variable $Z$ is $[0,\infty)$-valued. The RHS is the expectation of a non-negative random variable, hence $ZI_{(0,1)}(Z)$ equals zero almost surely. Explain why this implies $P[Z\in(0,1)]=0$. Continue in this way to prove that $P[Z\in(n,n+1)]=0$ for every $n=1,2,\ldots$.
Second hint: After you've proved that $Z$ takes only values $0, 1, 2,\ldots$, you can assert (1), using $(n,n+1]$ instead of $(n, n+1)$ on the LHS. If you know $Z$ is integer-valued, then $E[Z\cdot I_{(n,n+1]}(Z)] = (n+1) P(Z=n+1)$.