[proofplan] We prove the upper-tail estimate by comparing the indicator of the event $\{X\ge t\}$ with the exponential [random variable](/page/Random%20Variable) $e^{\lambda X}$ for each admissible $\lambda>0$. Taking expectations gives the usual Chernoff estimate with a free parameter $\lambda$. Since the exponential function is increasing, optimizing over $\lambda$ converts the infimum form into the Legendre-transform supremum form. The lower-tail estimate is the same argument with $\lambda<0$, where the sign of $\lambda$ reverses the pointwise comparison. [/proofplan]

[step:Bound the upper tail for one admissible positive parameter] Fix $t\in\mathbb R$ and fix $\lambda\in D_X\cap(0,\infty)$. Let \begin{align*} A_t:=\{\omega\in\Omega:X(\omega)\ge t\} \end{align*} denote the upper-tail event. Since $X$ is measurable and $[t,\infty)\in\mathcal B(\mathbb R)$, the set $A_t=X^{-1}([t,\infty))$ belongs to $\mathcal F$. For every $\omega\in A_t$, the inequality $\lambda>0$ gives $\lambda X(\omega)\ge \lambda t$, hence \begin{align*} 1\le \exp(\lambda X(\omega)-\lambda t). \end{align*} For every $\omega\notin A_t$, the indicator is $0$, so the same inequality holds after multiplying by $\mathbb 1_{A_t}$. Thus, pointwise on $\Omega$, \begin{align*} \mathbb 1_{A_t}\le \exp(-\lambda t)e^{\lambda X}. \end{align*} Taking expectations of both sides, using monotonicity of the [Lebesgue integral](/page/Lebesgue%20Integral) with respect to $\mathbb P$, and using $\lambda\in D_X$, we obtain \begin{align*} \mathbb P(X\ge t) =\mathbb E[\mathbb 1_{A_t}] \le \exp(-\lambda t)\mathbb E[e^{\lambda X}] =\exp(\Lambda_X(\lambda)-\lambda t). \end{align*} [/step]

[step:Optimize the upper-tail estimate over positive admissible parameters]The previous step gives, for every $\lambda\in D_X\cap(0,\infty)$, \begin{align*} \mathbb P(X\ge t)\le \exp(\Lambda_X(\lambda)-\lambda t). \end{align*} Taking the infimum over $\lambda\in D_X\cap(0,\infty)$ yields \begin{align*} \mathbb P(X\ge t) \le \inf_{\lambda\in D_X\cap(0,\infty)}\exp(\Lambda_X(\lambda)-\lambda t). \end{align*} Because $\exp:\mathbb R\to(0,\infty)$ is increasing, \begin{align*} \inf_{\lambda\in D_X\cap(0,\infty)}\exp(\Lambda_X(\lambda)-\lambda t) = \exp\left(\inf_{\lambda\in D_X\cap(0,\infty)}\{\Lambda_X(\lambda)-\lambda t\}\right). \end{align*} Finally, \begin{align*} \inf_{\lambda\in D_X\cap(0,\infty)}\{\Lambda_X(\lambda)-\lambda t\} = -\sup_{\lambda\in D_X\cap(0,\infty)}\{\lambda t-\Lambda_X(\lambda)\}. \end{align*} Therefore \begin{align*} \mathbb P(X\ge t)\le \exp\left(-\sup_{\lambda\in D_X\cap(0,\infty)}\{\lambda t-\Lambda_X(\lambda)\}\right). \end{align*}[/step]

[guided]We now turn the one-parameter estimate into the advertised Legendre-transform form. From the preceding step, every admissible positive parameter $\lambda\in D_X\cap(0,\infty)$ gives the valid bound \begin{align*} \mathbb P(X\ge t)\le \exp(\Lambda_X(\lambda)-\lambda t). \end{align*} Since the left-hand side does not depend on $\lambda$, it is bounded by the best of these right-hand sides: \begin{align*} \mathbb P(X\ge t) \le \inf_{\lambda\in D_X\cap(0,\infty)}\exp(\Lambda_X(\lambda)-\lambda t). \end{align*} The remaining point is purely order-theoretic. The exponential function $\exp:\mathbb R\to(0,\infty)$ is increasing, so taking the infimum after exponentiating is the same as exponentiating the infimum: \begin{align*} \inf_{\lambda\in D_X\cap(0,\infty)}\exp(\Lambda_X(\lambda)-\lambda t) = \exp\left(\inf_{\lambda\in D_X\cap(0,\infty)}\{\Lambda_X(\lambda)-\lambda t\}\right). \end{align*} The expression inside the exponential is then rewritten by factoring out a minus sign: \begin{align*} \inf_{\lambda\in D_X\cap(0,\infty)}\{\Lambda_X(\lambda)-\lambda t\} = -\sup_{\lambda\in D_X\cap(0,\infty)}\{\lambda t-\Lambda_X(\lambda)\}. \end{align*} Substituting this identity gives \begin{align*} \mathbb P(X\ge t)\le \exp\left(-\sup_{\lambda\in D_X\cap(0,\infty)}\{\lambda t-\Lambda_X(\lambda)\}\right). \end{align*} This is exactly the upper-tail [Chernoff bound](/theorems/6038) in Legendre transform form.[/guided]

[step:Repeat the pointwise comparison for the lower tail with negative parameters] Fix $\lambda\in D_X\cap(-\infty,0)$. Let \begin{align*} B_t:=\{\omega\in\Omega:X(\omega)\le t\} \end{align*} denote the lower-tail event. Since $X$ is measurable and $(-\infty,t]\in\mathcal B(\mathbb R)$, the set $B_t=X^{-1}((-\infty,t])$ belongs to $\mathcal F$. For every $\omega\in B_t$, the inequality $\lambda<0$ reverses order and gives $\lambda X(\omega)\ge \lambda t$. Therefore \begin{align*} 1\le \exp(\lambda X(\omega)-\lambda t). \end{align*} As before, this implies the pointwise inequality \begin{align*} \mathbb 1_{B_t}\le \exp(-\lambda t)e^{\lambda X}. \end{align*} Taking expectations and using $\lambda\in D_X$ gives \begin{align*} \mathbb P(X\le t) =\mathbb E[\mathbb 1_{B_t}] \le \exp(-\lambda t)\mathbb E[e^{\lambda X}] =\exp(\Lambda_X(\lambda)-\lambda t). \end{align*} Taking the infimum over $\lambda\in D_X\cap(-\infty,0)$ and using the same increasing-exponential and infimum-supremum identities as above, we obtain \begin{align*} \mathbb P(X\le t)\le \exp\left(-\sup_{\lambda\in D_X\cap(-\infty,0)}\{\lambda t-\Lambda_X(\lambda)\}\right). \end{align*} [/step]