I am having difficulty interpreting the following theorem from here as a probability statement:
Theorem.
For all $\delta$ such that $0 < \delta < 1/2$, with proability at least $1 - \delta$ the following is true $$\lvert \hat{R}_n(f) - {R}(f) \rvert \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}} \quad \forall f \in \mathcal{F} \tag{4.3}$$
Where $\hat{R}_n(f) = n^{-1} \sum^n_{i=1} \mathbb{I}(Y_i \neq f(X_i))$ is the empirical risk under a 0/1 loss function from $n$ training data points $(X_i, Y_i)$ drawn from an unknown distribution $p^*$, and $R(f) = \mathbb{E}_{(X, Y) \sim p^*}[\mathbb{I}(Y \neq f(X))]$ is the risk. $f$ is a linear classifier in a function class $\mathcal{F}$ that is finite i.e. $\lvert \mathcal{F} \rvert < \infty$.
Query.
If I were to rewrite this as a probability statement, would the qualifier $\forall f \in \mathcal{F}$ go inside or outside the probability statement?
That is, should the above theorem read:
$$\mathbb{P} \left( \forall f \in \mathcal{F} \lvert \hat{R}_n(f) - R(f) \rvert \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}}\right) \geq 1 - \delta \tag{A}$$
Or should it read:
$$\forall f \in \mathcal{F} , \mathbb{P} \left( \lvert \hat{R}_n(f) - R(f) \rvert \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}}\right) \geq 1 - \delta \tag{B}$$
Further context - proof of the theorem.
I will now detail each step of the proof, which is fairly standard in the statistics and statistical learning theory literature, together with my understanding.
Step 1 - Bound the deviation of the empirical risk from the risk for a single function $f \in \mathcal{F}$ using a concentration inequality.
For a single function $f \in \mathcal{F}$, as the empirical risk $\hat{R}(f)$ is the sample average of i.i.d. Bernoulli rvs, and the risk $R(f)$ is the theoretical expectation, we have that
$$\mathbb{P} \left( \lvert \hat{R}(f) - R(f) \rvert \geq \epsilon \right) \leq 2\exp(-2n\epsilon^2)$$
Step 2 - Bound the uniform deviation using the union bound.
My understanding is that we require the empirical risk to be a good estimate of the risk in the sense of being 'uniformly close' to the risk over the entire function class $\mathcal{F}$ . And that Hoeffding's inequality by itself is insufficient, because for a particular $f \in \mathcal{F}$, there will be a set of training samples $S_f$ such that $\hat{R}(f) - R(f) \leq \epsilon$ with probability $\mathbb{P}(S_f) \geq 1 - 2\exp(-2n\epsilon^2)$, and the crux is that in general, this set $S_f$ will be different for different functions in $\mathcal{F}$.
We define the event $E_f = \{\lvert \hat{R}(f) - R(f) \rvert \geq \epsilon\}$ and as the function class $\mathcal{F}$ is finite, apply the union bound over $E_f$:
$$\begin{align} \mathbb{P} \left( \sup_{f \in \mathcal{F}} \lvert \hat{R}_n(f) - R(f) \rvert \geq \epsilon \right) &= \mathbb{P}\left( \bigcup_{f \in \mathcal{F}} \{ \lvert \hat{R}_n(f) - R(f) \rvert \geq \epsilon\} \right) \\ &\leq \sum_{f \in \mathcal{F}} \mathbb{P} \left(\lvert \hat{R}_n(f) - R(f) \rvert \geq \epsilon\right) \\ &\leq \sum_{f \in \mathcal{F}} 2 \exp(-2n \epsilon^2) \\ &= 2 \lvert \mathcal{F} \rvert \exp(-2n \epsilon^2) \end{align}$$
Rewriting in terms of $\delta$, that is taking $\delta := 2 \lvert \mathcal{F} \rvert \exp(-2n \epsilon^2)$ and rearranging for $\epsilon$ to get
$$\epsilon = \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}}$$
Yields
$$\mathbb{P} \left( \sup_{f \in \mathcal{F}} \lvert \hat{R}_n(f) - R(f) \rvert \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}} \right) \geq 1 - \delta$$
Up to here, I am comfortable with the proof steps - and I read the above line as, "the probability that the maximum deviation between the empirical risk and risk over the entire function class $\mathcal{F}$ lies in $[-C_{\delta}, C_{\delta}]$ is at least $1 - \delta$.
My confusion.
However, where it gets unfamiliar is that the authors state that this proves $(4.3)$, but I am having difficulty trying to figure out whether this means that
$$\begin{align} &\mathbb{P} \left( \sup_{f \in \mathcal{F}} \lvert \hat{R}_n(f) - R(f) \rvert \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}} \right) \geq 1 - \delta \\ & \implies \mathbb{P} \left( \forall f \in \mathcal{F} \lvert \hat{R}_n(f) - R(f) \rvert \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}}\right) \geq 1 - \delta \end{align}$$
or whether it means that
$$\begin{align} &\mathbb{P} \left( \sup_{f \in \mathcal{F}} \lvert \hat{R}_n(f) - R(f) \rvert \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}} \right) \geq 1 - \delta \\ & \implies \forall f \in \mathcal{F} , \mathbb{P} \left( \lvert \hat{R}_n(f) - R(f) \rvert \leq \sqrt{\frac{\ln \lvert \mathcal{F} \rvert + \ln (2 / \delta)}{2n}}\right) \geq 1 - \delta \\ \end{align}$$
