Conditional Markov Inequality using Conditional Measures

Question

Setup
Let $(\Omega, \mathcal{A}, \mu)$ be a measure space and $\{A_{n}: \mu(A_{n})\lt\infty\}_{n\geqslant 1} \subset \mathcal{A} ~~$s.t. $A_{n} \uparrow \Omega$. With $d(\cdot,\cdot)$ being a metric (i.e. distance function), let $$ \widetilde{d}_{N}(f,f_{n}) = \int_{A_{N}} (1\wedge d(f(\omega),f_{n}(\omega)) \mu(d\omega) $$

Assume $\widetilde{d}_{N}(f,f_{n})\overset{n\rightarrow\infty}{\longrightarrow} 0$. Let $B\in\mathcal{A}$ with $\mu(B)\lt\infty$. Fix $\delta\gt 0$ and choose $N\in\mathbb{N}$ large enough that $\mu(B\setminus A_{N})\lt \delta$. Then, for $\epsilon \in (0,1)$ $$ \begin{align} \mu(B~\cap~\{d(f,f_{n})\ > \epsilon\}) \leq&~ \delta + \mu(A_{N}\cap\{d(f,f_{n})\ > \epsilon)\\ \leq&~ \delta + \epsilon^{-1}\widetilde{d}_{N}(f,f_{n}) \end{align} $$

Question
How does one show $$ \mu(A_{N}\cap\{d(f,f_{n}) > \epsilon \}) \leq \epsilon^{-1}\widetilde{d}_{N}(f,f_{n}) $$ Also any intuition behind what is going on would be welcome!

References:
Theorem 6.7 from A. Klenke, Probability Theory, Universitext, DOI 10.1007/978-1-4471-5361, © Springer-Verlag London 2014

Answer 1

Let $\widetilde{\mu}(B)=\dfrac{\mu(A_{N}\cap B)}{\mu(A_{N})}$ and $\epsilon\in (0,1)$. It is clear that $\widetilde{\mu}$ is a (conditional) probability measure, so we may apply Markov's inequality with the random variable $|X| = |1\wedge d(f,f_{n})| = 1\wedge d(f,f_{n})$: $$ \begin{align} \dfrac{\mu(A_{N}\cap \{\omega : d(f(\omega),f_{n}(\omega))\ > \epsilon\})}{\mu(A_{N})} =&~ \widetilde{\mu}(\{\omega : d(f(\omega),f_{n}(\omega))\ > \epsilon\})\\ =&~\widetilde{\mu}(\{\omega : 1\wedge d(f(\omega),f_{n}(\omega))\ > \epsilon\})\\[-1em]~\\[-1em] =&~\widetilde{\mu}(\{\omega : |X(\omega)| > \epsilon\})\\[-1em]~\\[-1em] \leq&~ \dfrac{\widetilde{E}(|X|)}{\epsilon} ~~\ast \text{Markov's Inequality} \\[-1em]~\\[-1em] =&~ \dfrac{\widetilde{E}(1\wedge d(f,f_{n}))}{\epsilon} \\[-1em]~\\[-1em] =&~ \dfrac{\int_{\Omega}1\wedge d(f(\omega),f_{n}(\omega))~\widetilde{\mu}(d\omega)}{\epsilon} \tag{1}\label{eq1} \\[-1em]~\\[-1em] =&~ \dfrac{\int_{A_{N}}1\wedge d(f(\omega),f_{n}(\omega))~\mu(d\omega)}{\mu(A_{N})\cdot\epsilon} \tag{2}\label{eq2} \\[-1em]~\\[-1em] =&~ \dfrac{\widetilde{d}_{N}(f,f_{n})}{\mu(A_{N})\cdot\epsilon} \end{align} $$

Then just multiply both sides of the inequality by $\mu(A_{N})$ and we have the result. I'm having doubts about the change from integration w.r.t $~\widetilde{\mu}$ on $\Omega$ in $\eqref{eq1}$ to integration w.r.t. $~\mu$ on $A_{N}$ in $\eqref{eq2}$.