[proofplan]
The proof is the exponential form of Markov's argument. For a positive parameter $\lambda$, the event $\{X\geq t\}$ forces $e^{\lambda X}\geq e^{\lambda t}$, so its indicator is bounded by $e^{-\lambda t}e^{\lambda X}$; taking expectations gives an upper-tail estimate for each admissible $\lambda>0$. The lower-tail estimate is the same comparison with $\lambda<0$, where the inequality reverses before applying the increasing exponential function. Taking the infimum over all admissible parameters gives the two stated bounds.
[/proofplan]
[step:Bound the upper tail for each positive admissible parameter]
Fix $t\in\mathbb{R}$ and fix $\lambda\in D_X\cap(0,\infty)$. Define the event
\begin{align*}
A_\lambda:=\{\omega\in\Omega:X(\omega)\geq t\}\in\mathcal{F}.
\end{align*}
For every $\omega\in A_\lambda$, since $\lambda>0$, we have $\lambda X(\omega)\geq \lambda t$, hence $e^{\lambda X(\omega)}\geq e^{\lambda t}$. Therefore, for every $\omega\in\Omega$,
\begin{align*}
\mathbb{1}_{A_\lambda}(\omega)\leq e^{-\lambda t}e^{\lambda X(\omega)}.
\end{align*}
Both sides are non-negative random variables, and $e^{\lambda X}$ is integrable because $\lambda\in D_X$. By monotonicity of expectation,
\begin{align*}
\mathbb{P}(X\geq t)
=\mathbb{E}[\mathbb{1}_{A_\lambda}]
\leq e^{-\lambda t}\mathbb{E}[e^{\lambda X}]
=e^{-\lambda t}M_X(\lambda).
\end{align*}
[guided]
Fix $t\in\mathbb{R}$ and choose an admissible positive parameter $\lambda\in D_X\cap(0,\infty)$. The word admissible means exactly that $\mathbb{E}[e^{\lambda X}]<\infty$, so the moment generating function value $M_X(\lambda)$ is finite.
Define the upper-tail event
\begin{align*}
A_\lambda:=\{\omega\in\Omega:X(\omega)\geq t\}.
\end{align*}
Since $X$ is a real-valued [random variable](/page/Random%20Variable), $A_\lambda=X^{-1}([t,\infty))\in\mathcal{F}$, so its probability and indicator function are well-defined.
The key comparison is pointwise. If $\omega\in A_\lambda$, then $X(\omega)\geq t$. Multiplying by the positive number $\lambda$ preserves the inequality:
\begin{align*}
\lambda X(\omega)\geq \lambda t.
\end{align*}
The exponential function is increasing on $\mathbb{R}$, so
\begin{align*}
e^{\lambda X(\omega)}\geq e^{\lambda t}.
\end{align*}
Equivalently,
\begin{align*}
1\leq e^{-\lambda t}e^{\lambda X(\omega)}
\end{align*}
on $A_\lambda$. If $\omega\notin A_\lambda$, then $\mathbb{1}_{A_\lambda}(\omega)=0$, while $e^{-\lambda t}e^{\lambda X(\omega)}>0$. Hence the pointwise indicator bound holds on all of $\Omega$:
\begin{align*}
\mathbb{1}_{A_\lambda}(\omega)\leq e^{-\lambda t}e^{\lambda X(\omega)}.
\end{align*}
Now take expectations. The right-hand side is integrable because $\lambda\in D_X$, and both random variables are non-negative. Monotonicity of expectation gives
\begin{align*}
\mathbb{P}(X\geq t)
=\mathbb{E}[\mathbb{1}_{A_\lambda}]
\leq \mathbb{E}[e^{-\lambda t}e^{\lambda X}]
=e^{-\lambda t}\mathbb{E}[e^{\lambda X}]
=e^{-\lambda t}M_X(\lambda).
\end{align*}
This proves the upper-tail estimate for this particular admissible $\lambda>0$.
[/guided]
[/step]
[step:Take the infimum over positive admissible parameters]
The previous step shows that
\begin{align*}
\mathbb{P}(X\geq t)\leq e^{-\lambda t}M_X(\lambda)
\end{align*}
for every $\lambda\in D_X\cap(0,\infty)$. Therefore $\mathbb{P}(X\geq t)$ is a lower bound for the set of right-hand-side values, and hence
\begin{align*}
\mathbb{P}(X\geq t)\leq \inf_{\lambda\in D_X\cap(0,\infty)} e^{-\lambda t}M_X(\lambda).
\end{align*}
If $D_X\cap(0,\infty)=\varnothing$, the right-hand side is $+\infty$ by convention, and the inequality remains valid.
[/step]
[step:Bound the lower tail for each negative admissible parameter]
Fix $\lambda\in D_X\cap(-\infty,0)$. Define the event
\begin{align*}
B_\lambda:=\{\omega\in\Omega:X(\omega)\leq t\}\in\mathcal{F}.
\end{align*}
For every $\omega\in B_\lambda$, since $\lambda<0$, multiplying $X(\omega)\leq t$ by $\lambda$ gives $\lambda X(\omega)\geq \lambda t$. Applying the increasing exponential function gives $e^{\lambda X(\omega)}\geq e^{\lambda t}$. Thus, for every $\omega\in\Omega$,
\begin{align*}
\mathbb{1}_{B_\lambda}(\omega)\leq e^{-\lambda t}e^{\lambda X(\omega)}.
\end{align*}
Taking expectations and using $\lambda\in D_X$,
\begin{align*}
\mathbb{P}(X\leq t)
=\mathbb{E}[\mathbb{1}_{B_\lambda}]
\leq e^{-\lambda t}\mathbb{E}[e^{\lambda X}]
=e^{-\lambda t}M_X(\lambda).
\end{align*}
[/step]
[step:Take the infimum over negative admissible parameters]
The previous step gives
\begin{align*}
\mathbb{P}(X\leq t)\leq e^{-\lambda t}M_X(\lambda)
\end{align*}
for every $\lambda\in D_X\cap(-\infty,0)$. Taking the infimum over all such $\lambda$ yields
\begin{align*}
\mathbb{P}(X\leq t)\leq \inf_{\lambda\in D_X\cap(-\infty,0)} e^{-\lambda t}M_X(\lambda).
\end{align*}
If $D_X\cap(-\infty,0)=\varnothing$, the right-hand side is $+\infty$ by convention. This proves both claimed Chernoff bounds.
[/step]
