Fix hypercube $\{ 0, 1 \}^n$. Let $A \subseteq \{ 0, 1 \}^n$ of size $|A| \geq 2^{n - 1}$. Then it is claimed $\forall t > 0$, $$ |\{ \nu \in \{ 0, 1 \}^n: \mathrm{dist}_{\{ 0, 1 \}^n}(\nu, A) > t \}| \leq 2\exp\left( -\frac{t^2}{16n} \right). $$ I am trying to follow this proof:
Fix $f(\nu) := \mathrm{dist}_{\{0, 1\}^n}(\nu, A)$-is coordinate-wise $1$-Lipschitz. Hence if $X$ is uniform, random vertex of $\{ 0, 1 \}^n$, then through bounded difference inequality $$ \mathbb{P}\{ |f(X) - \mathbb{E} f(X)| \geq t \} \leq 2\exp\left( -\frac{t^2}{4n} \right). $$ Remains to note: $\mathbb{E} f(X) = O(\sqrt{n})$.
I am not very sure if I understood this proof: I follow everything until the last sentence. How do we know $\mathbb{E} f(X) = O(\sqrt{n})$ and how does knowing $\mathbb{E}f(X) = O(\sqrt{n})$ help us to obtain the claim?