[proofplan] The proof is the standard exponential moment argument. For a fixed admissible $\lambda>0$, we compare the event $\{X\ge a\}$ with the nonnegative [random variable](/page/Random%20Variable) $e^{\lambda X}$, using the monotonicity of the exponential function. Taking expectations of the resulting pointwise inequality gives the one-parameter bound, and then taking the infimum over all admissible $\lambda$ gives the optimized form. [/proofplan] [step:Compare the threshold event with the exponential random variable] Fix $a \in \mathbb R$ and $\lambda \in D_X$. Define the event \begin{align*} A_a := \{\omega \in \Omega : X(\omega) \ge a\} \in \mathcal F \end{align*} and define the nonnegative random variable $Y_\lambda:\Omega \to [0,\infty)$ by $Y_\lambda(\omega) := e^{\lambda X(\omega)}$ for every $\omega \in \Omega$. The measurability of $Y_\lambda$ follows from the measurability of $X$ and continuity of $t\mapsto e^{\lambda t}$. Since $\lambda \in D_X$, $Y_\lambda$ is integrable. Because $\lambda>0$, the function $t\mapsto e^{\lambda t}$ is increasing on $\mathbb R$. Hence, for every $\omega \in A_a$, \begin{align*} Y_\lambda(\omega)=e^{\lambda X(\omega)} \ge e^{\lambda a}. \end{align*} Equivalently, for every $\omega \in \Omega$, \begin{align*} e^{\lambda a}\mathbb{1}_{A_a}(\omega) \le Y_\lambda(\omega). \end{align*} [guided] Fix $a \in \mathbb R$ and choose an admissible exponential parameter $\lambda \in D_X$. The admissibility means precisely that \begin{align*} \mathbb E[e^{\lambda X}] < \infty, \end{align*} so the exponential random variable we are about to use is integrable. Define the event \begin{align*} A_a := \{\omega \in \Omega : X(\omega) \ge a\}. \end{align*} Since $X:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable and $[a,\infty)$ is a Borel subset of $\mathbb R$, we have \begin{align*} A_a = X^{-1}([a,\infty)) \in \mathcal F. \end{align*} Now define the nonnegative random variable $Y_\lambda:\Omega \to [0,\infty)$ by $Y_\lambda(\omega) := e^{\lambda X(\omega)}$ for every $\omega \in \Omega$. This is measurable because it is the composition of the measurable map $X$ with the continuous map $t\mapsto e^{\lambda t}$. It is integrable because $\lambda \in D_X$. The reason for exponentiating is that the event $X\ge a$ becomes a multiplicative lower bound. Since $\lambda>0$, the map $t\mapsto e^{\lambda t}$ is increasing. Therefore, if $\omega\in A_a$, then $X(\omega)\ge a$, and hence \begin{align*} Y_\lambda(\omega)=e^{\lambda X(\omega)} \ge e^{\lambda a}. \end{align*} If $\omega\notin A_a$, then $\mathbb{1}_{A_a}(\omega)=0$, so \begin{align*} e^{\lambda a}\mathbb{1}_{A_a}(\omega)=0 \le Y_\lambda(\omega). \end{align*} Combining the two cases gives the pointwise inequality \begin{align*} e^{\lambda a}\mathbb{1}_{A_a}(\omega) \le Y_\lambda(\omega) \end{align*} for every $\omega \in \Omega$. [/guided] [/step] [step:Take expectations to obtain the one-parameter bound] Taking expectations in the pointwise inequality and using monotonicity of expectation gives \begin{align*} e^{\lambda a}\mathbb P(A_a) = \mathbb E[e^{\lambda a}\mathbb{1}_{A_a}] \le \mathbb E[Y_\lambda] = \mathbb E[e^{\lambda X}]. \end{align*} Since $e^{\lambda a}>0$, division by $e^{\lambda a}$ yields \begin{align*} \mathbb P(X\ge a) = \mathbb P(A_a) \le e^{-\lambda a}\mathbb E[e^{\lambda X}]. \end{align*} By the definition of $\Lambda_X(\lambda)$, \begin{align*} e^{-\lambda a}\mathbb E[e^{\lambda X}] = \exp(-\lambda a)\exp(\Lambda_X(\lambda)) = \exp(\Lambda_X(\lambda)-\lambda a). \end{align*} Therefore \begin{align*} \mathbb P(X\ge a) \le \exp(\Lambda_X(\lambda)-\lambda a). \end{align*} [/step] [step:Optimize over all admissible exponential parameters] The preceding estimate holds for every $\lambda \in D_X$. Hence, if $D_X$ is nonempty, then for every $a\in\mathbb R$, \begin{align*} \mathbb P(X\ge a) \le \inf_{\lambda \in D_X}\exp(\Lambda_X(\lambda)-\lambda a). \end{align*} This is exactly the optimized Chernoff bound. [/step]