[proofplan] The proof converts a tail event for $X$ into a tail event for the non-negative [random variable](/page/Random%20Variable) $e^{\lambda X}$. For $\lambda>0$, the exponential map is increasing, so $X\ge t$ forces $e^{\lambda X}\ge e^{\lambda t}$; for $\lambda<0$, it is decreasing, so $X\le t$ forces the same exponential lower bound. In both cases we use the pointwise indicator inequality $\mathbb 1_A\le e^{-\lambda t}e^{\lambda X}$ and integrate with respect to $\mathbb P$. [/proofplan]

[step:Convert the upper-tail event into an indicator bound]Assume $\lambda>0$ and $\mathbb E[e^{\lambda X}]<\infty$. Define the continuous map $h_\lambda:\mathbb R\to(0,\infty)$ by $h_\lambda(x)=e^{\lambda x}$ for every $x\in\mathbb R$. Let $Y_\lambda:\Omega\to(0,\infty)$ be the random variable $Y_\lambda=h_\lambda\circ X$, so $Y_\lambda(\omega)=e^{\lambda X(\omega)}$ for every $\omega\in\Omega$. Since $h_\lambda$ is increasing when $\lambda>0$, the event \begin{align*} A:=\{\omega\in\Omega:X(\omega)\ge t\} \end{align*} satisfies $Y_\lambda(\omega)\ge e^{\lambda t}$ for every $\omega\in A$. Hence, for every $\omega\in\Omega$, \begin{align*} \mathbb 1_A(\omega)\le e^{-\lambda t}Y_\lambda(\omega). \end{align*}[/step]

[guided]Assume $\lambda>0$ and $\mathbb E[e^{\lambda X}]<\infty$. We first isolate the non-negative random variable to which the argument will be applied. Define the map $h_\lambda:\mathbb R\to(0,\infty)$ by $h_\lambda(x)=e^{\lambda x}$ for every $x\in\mathbb R$. The map $h_\lambda$ is continuous, hence Borel measurable. Since $X:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable, the composition $Y_\lambda:\Omega\to(0,\infty)$, defined by $Y_\lambda(\omega)=h_\lambda(X(\omega))=e^{\lambda X(\omega)}$ for every $\omega\in\Omega$, is a non-negative random variable. The point of introducing $Y_\lambda$ is that the event $X\ge t$ can be compared pointwise with a lower bound on $Y_\lambda$. Define \begin{align*} A:=\{\omega\in\Omega:X(\omega)\ge t\}. \end{align*} Because $X$ is measurable and $[t,\infty)\in\mathcal B(\mathbb R)$, the set $A=X^{-1}([t,\infty))$ belongs to $\mathcal F$. Since $\lambda>0$, the function $h_\lambda$ is increasing. Therefore, if $\omega\in A$, then $X(\omega)\ge t$, and so \begin{align*} Y_\lambda(\omega)=e^{\lambda X(\omega)}\ge e^{\lambda t}. \end{align*} Multiplying by the positive number $e^{-\lambda t}$ gives \begin{align*} 1\le e^{-\lambda t}Y_\lambda(\omega) \end{align*} for every $\omega\in A$. If $\omega\notin A$, then $\mathbb 1_A(\omega)=0$, while $e^{-\lambda t}Y_\lambda(\omega)>0$. Combining the two cases gives the pointwise inequality \begin{align*} \mathbb 1_A(\omega)\le e^{-\lambda t}Y_\lambda(\omega) \end{align*} for every $\omega\in\Omega$.[/guided]

[step:Integrate the upper-tail indicator bound] Since $Y_\lambda\ge0$ and $\mathbb E[Y_\lambda]=\mathbb E[e^{\lambda X}]<\infty$, both sides of the pointwise inequality are integrable. Integrating with respect to $\mathbb P$ gives \begin{align*} \mathbb P(X\ge t)=\mathbb P(A)=\int_\Omega \mathbb 1_A(\omega)\,d\mathbb P(\omega). \end{align*} The pointwise inequality and monotonicity of the integral give \begin{align*} \int_\Omega \mathbb 1_A(\omega)\,d\mathbb P(\omega)\le e^{-\lambda t}\int_\Omega Y_\lambda(\omega)\,d\mathbb P(\omega). \end{align*} By the definition of $Y_\lambda$ and of expectation, \begin{align*} e^{-\lambda t}\int_\Omega Y_\lambda(\omega)\,d\mathbb P(\omega)=e^{-\lambda t}\mathbb E[e^{\lambda X}]. \end{align*} This proves the upper-tail bound. [/step]

[step:Convert the lower-tail event into the same indicator bound] Assume $\lambda<0$ and $\mathbb E[e^{\lambda X}]<\infty$. Define the map $h_\lambda:\mathbb R\to(0,\infty)$ by $h_\lambda(x)=e^{\lambda x}$ for every $x\in\mathbb R$, and let $Y_\lambda:\Omega\to(0,\infty)$ be given by $Y_\lambda=h_\lambda\circ X$. Since $h_\lambda$ is decreasing when $\lambda<0$, the event \begin{align*} B:=\{\omega\in\Omega:X(\omega)\le t\} \end{align*} satisfies $Y_\lambda(\omega)\ge e^{\lambda t}$ for every $\omega\in B$. Therefore, for every $\omega\in\Omega$, \begin{align*} \mathbb 1_B(\omega)\le e^{-\lambda t}Y_\lambda(\omega). \end{align*} [/step]

[step:Integrate the lower-tail indicator bound] Since $Y_\lambda\ge0$ and $\mathbb E[Y_\lambda]=\mathbb E[e^{\lambda X}]<\infty$, integrating the pointwise inequality over $\Omega$ with respect to $\mathbb P$ yields \begin{align*} \mathbb P(X\le t)=\mathbb P(B)=\int_\Omega \mathbb 1_B(\omega)\,d\mathbb P(\omega). \end{align*} The pointwise inequality and monotonicity of the integral give \begin{align*} \int_\Omega \mathbb 1_B(\omega)\,d\mathbb P(\omega)\le e^{-\lambda t}\int_\Omega Y_\lambda(\omega)\,d\mathbb P(\omega). \end{align*} By the definition of $Y_\lambda$ and of expectation, \begin{align*} e^{-\lambda t}\int_\Omega Y_\lambda(\omega)\,d\mathbb P(\omega)=e^{-\lambda t}\mathbb E[e^{\lambda X}]. \end{align*} This proves the lower-tail bound and completes the proof. [/step]