Probabilistic solution to the Dirichlet problem with Dynkin formula

Question

I have two small questions to the probabilistic solution to the Dirichlet problem using Dynkin's formula. Here is how we introduced it and I marked the two equalities I'm referring to.

1. Why is $\int_0^{\infty} P_x(\tau_G > t) dt \leq \sum_{n=0}^{\infty} P_x(\tau_G >n)$?
2. How do we use the tower property for the second equation? $P_x(\tau_G > n-1, \tau_G(n-1)>n)=E_x[1_{\{\tau_G>n-1\}}P_x(\tau_G(n-1)>n)|\mathcal{F}_n-1)]$

Answer 1

1. This follows from $$ \int_0^{\infty} P_x(\tau_G>t)dt=\sum_{n=0}^{\infty} \int_n^{n+1}P_x(\tau_G>t)dt\leq \sum_{n=0}^{\infty} P_x(\tau_G>n)\int_n^{n+1}dt=\sum_{n=0}^{\infty} P_x(\tau_G>n), $$ using the fact that $t\mapsto P_x(\tau_G>t)$ is nonincreasing.

2. We have \begin{align} P_x(\tau_G>n-1,\tau_G(n-1)>n)&=\mathbb{E}[\mathbf{1}(\tau_G>n-1)\mathbf{1}(\tau_G(n-1)>n)]\\ &=\mathbb{E}[\mathbb{E}[\mathbf{1}(\tau_G>n-1)\mathbf{1}(\tau_G(n-1)>n)\vert \mathcal{F}_{n-1}]]\\ &=\mathbb{E}[\mathbf{1}(\tau_G>n-1)\mathbb{E}[\mathbf{1}(\tau_G(n-1)>n)\vert \mathcal{F}_{n-1}]]\\ &=\mathbb{E}[\mathbf{1}(\tau_G>n-1)\Pr(\tau_G(n-1)>n\vert \mathcal{F}_{n-1})], \end{align} where the second equality is the tower property and the third is from $\mathcal{F}_{n-1}$-measurability of the first indicator.