[proofplan] We use Talagrand's convex distance inequality in its exponential-moment form for product probability spaces. The complement of the convex-distance enlargement is exactly the event on which the convex distance is larger than $r$. A one-line Markov estimate applied to the non-negative [random variable](/page/Random%20Variable) $\exp(d_T(X,A)^2/4)$ gives the desired tail bound, and the final display follows by substituting $\mathbb P(X \in A) \ge 1/2$. [/proofplan] [step:Apply Talagrand's convex distance inequality to the set $A$] Let $Z:\Omega\to[0,\infty)$ be the map defined by $Z(\omega)=d_T(X(\omega),A)$. The theorem statement assumes that the map $x\mapsto d_T(x,A)$ is $\mathcal E$-measurable, and $X$ is $(\mathcal F,\mathcal E)$-measurable, so $Z$ is an $\mathcal F$-measurable non-negative random variable. Since $X_1,\dots,X_n$ are independent, the law of $X$ is the product measure of the laws of the coordinates. Since $A \in \mathcal E$, $\mathbb P(X \in A)>0$, and $x\mapsto d_T(x,A)$ is measurable, the exponential-moment form of Talagrand's convex distance inequality for product spaces applies to $A$ and gives \begin{align*} \mathbb E\left[\exp\left(\frac{Z^2}{4}\right)\right] \le \frac{1}{\mathbb P(X \in A)}. \end{align*} Here we are citing a result not yet in the wiki: Talagrand's convex distance inequality. [guided] Define $Z:\Omega\to[0,\infty)$ by $Z(\omega)=d_T(X(\omega),A)$. This is the quantity whose tail we need to estimate, because $A_r^{\mathrm T}$ was defined by the condition $d_T(x,A)\le r$. Before taking expectations, we verify measurability: the theorem statement assumes that $x\mapsto d_T(x,A)$ is $\mathcal E$-measurable, and the random vector $X:\Omega\to E$ is $(\mathcal F,\mathcal E)$-measurable, so their composition $Z$ is an $\mathcal F$-measurable non-negative random variable. We now invoke Talagrand's convex distance inequality for product probability spaces. Its hypotheses are the product-space hypotheses in the statement together with measurability of the distance function: the coordinates $X_1,\dots,X_n$ are independent, so the law of the vector-valued map $X:\Omega \to E$ is the product of the coordinate laws; the set $A$ is measurable with positive probability; and $x\mapsto d_T(x,A)$ is measurable, so the exponential moment is well-defined. Therefore the inequality applies and yields the exponential-moment estimate \begin{align*} \mathbb E\left[\exp\left(\frac{Z^2}{4}\right)\right] \le \frac{1}{\mathbb P(X \in A)}. \end{align*} This is the only deep input in the proof. The remaining argument is the standard passage from an exponential-moment bound to a tail bound. Here we are citing a result not yet in the wiki: Talagrand's convex distance inequality. [/guided] [/step] [step:Convert the exponential moment bound into a tail bound] Fix $r \ge 0$. Since $Z$ is measurable, the event $\{Z>r\}$ belongs to $\mathcal F$. By the definition of $A_r^{\mathrm T}$, we have $X \notin A_r^{\mathrm T} \iff d_T(X,A)>r \iff Z>r$. Hence \begin{align*} \mathbb P(X \notin A_r^{\mathrm T})=\mathbb P(Z>r). \end{align*} On the event $\{Z>r\}$, the monotonicity of the exponential function gives \begin{align*} \exp\left(\frac{Z^2}{4}\right)>\exp\left(\frac{r^2}{4}\right). \end{align*} Therefore, \begin{align*} \exp\left(\frac{r^2}{4}\right)\mathbb P(Z>r) \le \mathbb E\left[\exp\left(\frac{Z^2}{4}\right)\right]. \end{align*} Combining this with the preceding step gives \begin{align*} \mathbb P(X \notin A_r^{\mathrm T}) \le \frac{1}{\mathbb P(X \in A)}\exp\left(-\frac{r^2}{4}\right). \end{align*} [/step] [step:Specialize the bound when $A$ has probability at least one half] If $\mathbb P(X \in A)\ge \frac{1}{2}$, then \begin{align*} \frac{1}{\mathbb P(X \in A)} \le 2. \end{align*} Substituting this into the previous estimate gives, for every $r \ge 0$, \begin{align*} \mathbb P(X \notin A_r^{\mathrm T}) \le 2\exp\left(-\frac{r^2}{4}\right). \end{align*} This is the stated particular case. [/step]