Let $I_n=\{0,1\}^n$ be the Boolean cube, the set of all $2^n$ sequences of length $n$ of $0’s$ and $1’s$. $I_n$ is made a metric space by introducing the Hamming distance between two points $x = (\xi_1, \dots , \xi_n)$ and $y = (\eta_1, \dots , \eta_n)$ as the number of coordinates where $x,y$ disagree: $$dist(x, y) = \lvert i: \xi_i \neq \eta_i \rvert.$$
I would like to proof that for $f : I_n \rightarrow \mathbb{R}$ with the bounded difference $$\lvert f(x) - f(y) \rvert \leq dist(x,y) \ \ \forall x,y \in I_n$$ and the average value $a_n$ of $f$ on $I_n$ being $a_f = \int_{I_n} f d\mu_n$ that \begin{equation} \mu_n \left \{ x \in I_n : \lvert f(x) -a_f \rvert \geq \epsilon \sqrt{n} \right \} \leq 2 e^{(-2\epsilon{^2})} \end{equation} with $\epsilon \geq 0$. So far I've found the formula above and a corresponding proof in relation to the chromatic number of random graphs, but nothing specific to the Hamming Cube. Does anybody know how I can specifically proof the measure concentration inequality in relation to this special type of graph?
I'm an EE undergrad who just started dabbling with mathematics.
Edit: sry Latex was weird
This is just an application of McDiarmid's inequality.