How can I perform Union operation into two events? Assume there are two events A and B such as $A := \frac{1}{n} \sum_{i=1}^{n}\left(f\left(x_{i}\right)-\mathbb{E}[f]\right) z_{i} \geq \frac{\epsilon}{8}$
and $B := \exists f \in \mathcal{F}: \frac{1}{n} \sum_{i=1}^{n} \mathbb{E}[f] z_{i} \geq \frac{\epsilon}{8}$
and P(A) and P(B) is given to me! I need to Calculate the Upperbound of the P(C) where C is defiend as
$$ C:= \mathbb{P}\left(\exists f \in \mathcal{F}: \frac{1}{n} \sum_{i=1}^{n} f\left(x_{i}\right) z_{i} \geq \frac{\epsilon}{4}\right)$$
I have used the upper bound $P(A \cup B) \leq P(A) + P(B)$
Further  I have reached upto
Given to me := \begin{align}
   \mathbb{P}\left (\frac{1}{n}\left| \sum_{i=1}^n (f(x_i) - \mathbb{E}[f])z_i\right| \geq \frac{\epsilon}{8} \right )  \leq 2\exp\left (-\frac{\epsilon^2nd}{9^4cL^2} \right)
... (1) \end{align}
\begin{align}
    \mathbb{P}\left(\exists f \in \mathcal{F}: \frac{1}{n} \sum_{i=1}^{n} \mathbb{E}[f] z_{i} \geq \frac{\epsilon}{8}\right) \leq \mathbb{P}\left(\left|\frac{1}{n} \sum_{i=1}^{n} z_{i}\right| \geq \frac{\epsilon}{8}\right) 
...(2)\end{align}
\begin{align}
\mathbb{P}(A \cup B)\\
&= \mathbb{P} \left( \underbrace{ \left( \frac{1}{n} \left|\sum_{i=1}^{n}\left(f\left(x_{i}\right)-\mathbb{E}[f]\right) z_{i}\right| \geq \frac{\epsilon}{8} \right)}_{A} \bigcup \underbrace{\left( \exists f \in \mathcal{F}: \frac{1}{n} \sum_{i=1}^{n} \mathbb{E}[f] z_{i} \geq \frac{\epsilon}{8}\right)}_{B}\right)\\
&\leq \mathbb{P}\left (  \frac{1}{n}\left| \sum_{i=1}^n (f(x_i) - \mathbb{E}[f])z_i\right| \geq \frac{\epsilon}{8} \right ) + \mathbb{P}\left(\exists f \in \mathcal{F}: \frac{1}{n} \sum_{i=1}^{n} \mathbb{E}[f] z_{i} \geq \frac{\epsilon}{8}\right)\\
&\leq 2 \exp\left (-\frac{\epsilon^2nd}{9^4cL^2} \right) + \mathbb{P}\left(\left|\frac{1}{n} \sum_{i=1}^{n} z_{i}\right| \geq \frac{\epsilon}{8}\right)  && \text{(Using      Inequality 1 and 2)}\\
&\leq 2 \exp\left (-\frac{\epsilon^2nd}{9^4cL^2} \right) + 2 \exp \left(\frac{-n \epsilon^{2}}{ 8^{3}}\right) && \text{(By Hoeffding's inequality)}
\end{align}
Can I proof LHS as with event C. I am not getting clue? Can anyone please help!
My question is := can I reduce $P(A \cup B) = P(C)$ if yes then how?
N.B > I am trying to understand the  proof of Theorem 2(Page -$7$) in the paper "A Universal Law of Robustness via isoperimetry" by Bubeck and Sellke.
 A: For two events $E,F$ one always has that if "$E \implies F$" then $P(E) \leq P(F)$. To see this note
\begin{align}
    P(F) &= P(E \text{ and } F) + P(F \text{ and not } E) \\
    &= P(F|E)P(E) + P(F \text{ and not } E) \\
    &= 1 P(E) + P(F \text{ and not } E) \\ 
    &\geq P(E)
\end{align}
where we have used that $P(F|E) = 1$ since $E \implies F$, and also $P(F \text{ and not } E) \geq 0$.
If we apply this with $E = C$ and $F = A \cup B$ then we will have $P(C) \leq P(A \cup B)$ which is what you need above. So all we need to do is check that $C \implies A \cup B$.
Suppose that $C$ holds and let $f \in \mathcal{F}$ be a function such that $\frac{1}{n}\sum_{i=1}^n f(x_i)z_i \geq \epsilon /4$. Suppose for contradiction that neither $A$ nor $B$ occurs so that we have both
\begin{align}
\frac{1}{n}\left | \sum_{i=1}^n (f(x_i) - \mathbb{E}f)z_i \right | &< \epsilon / 8 \\ 
\frac{1}{n}\sum_{i=1}^n\mathbb{E}[f]z_i &< \epsilon /8.
\end{align}
By the first inequality we have
\begin{align}
\frac{1}{n}\left ( \sum_{i=1}^n (f(x_i) - \mathbb{E}f)z_i \right ) &< \frac{\epsilon}{8} \\
\implies \frac{1}{n}\sum_{i=1}^n f(x_i)z_i&< \frac{\epsilon}{8} + \frac{1}{n}\sum_{i=1}^n \mathbb{E}[f]z_i
\end{align}
and by the second we have
\begin{align}
\frac{1}{n}\sum_{i=1}^n f(x_i)z_i &< \frac{\epsilon}{8} + \frac{1}{n}\sum_{i=1}^n \mathbb{E}[f]z_i \\
&< \frac{\epsilon}{8} + \frac{\epsilon}{8} = \frac{\epsilon}{4}
\end{align}
but this is a contradiction since we assumed that $\frac{1}{n}\sum_{i=1}^n f(x_i)z_i \geq \frac{\epsilon}{4}$.
In particular this implies that when $C$ holds at least one of $A$ or $B$ must hold, which is to say $A \cup B$ holds. This confirms that indeed $C \implies A \cup B$ and completes the proof.
A: I have answered here : Union Bound of two events?
But I am not getting How that |F| is coming in the RHS.
