I'm trying to understand how inequality $(5.3)$ follows from the inequality above it in this paper. The explanation given in the paper is that $(5.3)$ just follows from the previous inequality, so, I think the reason is trivial but I still don't understand.
I don't think a lot of background on what different random variables mean is necessary. Nevertheless, $G_n^\ast$ is a random graph, $X_t$ is an ergodic simple random walk on $G_n^\ast$ and $\pi$ is its stationary distribution. TV stands for total-variation (L-1) distance. Since $G_n^\ast$ is random, the distance $||P_x(X_{t+s}\in \cdot\vert G_n^\ast)-\pi||_{\rm{TV}}$ is random.
Inequality above $(5.3)$ is: $$\mathbb{P}(G_n^\ast\in \mathcal{G}\vert \mathcal{F}_{i-1})\geq 1-\varepsilon$$
Inequality $(5.3)$ is: $$ \mathbb{P}\left(||\mathbb{P}_x(X_{t+s}\in \cdot \vert G_n^\ast)-\pi ||_{\rm{TV}} = 1\{G_n^\ast\in \mathcal{G}\}||\mathbb{P}_x(X_{t+s}\in \cdot \vert G_n^\ast)-\pi ||_{\rm{TV}}\vert \mathcal{F}_{i-1}\right) \geq 1-\varepsilon $$
Any ideas?
Letting $d_x(t+s)$ denote the random TV term, you have $$P(d_x(t+s) = \mathbf{1}\{G_n^* \in \mathcal{G}\} d_x(t+s) \mid \mathcal{F}_{i-1}) \ge P(1 = \mathbf{1}\{G_n^* \in \mathcal{G}\} \mid \mathcal{F}_{i-1}) \ge 1-\epsilon,$$ because $1 = \mathbf{1}\{G_n^* \in \mathcal{G}\}$ implies $d_x(t+s) = \mathbf{1}\{G_n^* \in \mathcal{G}\} d_x(t+s)$. It is not equality because the latter equality could hold if $d_x(t+s)=0$ while $1 \ne \mathbf{1}\{G_n^* \in \mathcal{G}\}$.