How does "concentration inequalities" for conditional probability works? The conditional probability I am considering is
$$\mathbb{P}\left\{ \left| v'(A) - v_n(A)\right| < \frac{\epsilon}{2} \mid Z_1,\cdots,Z_n \right\}$$
where $v(A) = \mathbb{P}\left(Z_1 \in A\right)$ and $v_n'(A) = \frac{1}{n}\sum_{i=1}^n I(Z'_i \in A)$ (so-called empirical measure). In total, we have $2n$ iid random variables: $Z_1,\cdots,Z_n ,Z_1', \cdots,Z_n'$.
The derivation I have claims that "bounded by Chebyshev's inequality", we have
\begin{align*} \mathbb{P}\left\{ \left| v_n'(A) - v(A)\right| < \frac{\epsilon}{2} \mid Z_1,\cdots,Z_n \right\} \ge 1 - \frac{v(A)(1 - v(A))}{n\epsilon^2/4} \end{align*}
What I understand: I recognize that $\mathbb{E}\left[ v'_n(A)\right]=\frac{1}{n}\sum_{i=1}^n\mathbb{E}\left[ I(Z_i' \in A) \right] = \mathbb{P}\left(Z_1\in A\right) = v(A)$. Also, I understand that $\text{Var}\left[ v'_n(A)\right] = v(A)(1-v(A))$
What I don't understand: How Chebyshev's inequality can/should be applied to a conditional probability? First, in the usual setting, the probability we want to bound is really a number, but a conditional probability is a random variable. So what does the upper bound here exactly means? Bounded almost everywhere? I don't quite understand how conditional affects how things work out here...