[proofplan]
We decompose each [random variable](/page/Random%20Variable) into its truncated part and its large-value tail. This gives a pointwise identity for the full sum as the sum of a truncated contribution and a tail contribution. If the full sum has absolute value at least $t$, then the elementary inequality $|a+b|\leq |a|+|b|$ forces either the truncated contribution or the tail contribution to have size at least $t/2$. Finally, the absolute value of the tail contribution is bounded pointwise by the sum of the absolute tail magnitudes, and taking probabilities gives the desired inequality.
[/proofplan]
[step:Split the sum into its truncated and tail contributions]
For each $i \in \{1,\dots,n\}$, define the tail random variable $Y_i:\Omega \to \mathbb{R}$ by
\begin{align*}
Y_i(\omega) = X_i(\omega)\,\mathbb{1}_{\{|X_i|>L\}}(\omega),
\qquad \omega \in \Omega.
\end{align*}
Since
\begin{align*}
\mathbb{1}_{\{|X_i|\leq L\}}(\omega)+\mathbb{1}_{\{|X_i|>L\}}(\omega)=1
\end{align*}
for every $\omega \in \Omega$, we have the pointwise identity
\begin{align*}
X_i(\omega)=X_{i,L}(\omega)+Y_i(\omega).
\end{align*}
Define the random variables $S,A,B,C:\Omega \to \mathbb{R}$ as follows. Let
\begin{align*}
S(\omega)=\sum_{i=1}^n X_i(\omega).
\end{align*}
Let
\begin{align*}
A(\omega)=\sum_{i=1}^n X_{i,L}(\omega).
\end{align*}
Let
\begin{align*}
B(\omega)=\sum_{i=1}^n Y_i(\omega).
\end{align*}
Let
\begin{align*}
C(\omega)=\sum_{i=1}^n |X_i(\omega)|\,\mathbb{1}_{\{|X_i|>L\}}(\omega).
\end{align*}
Then $S=A+B$ pointwise on $\Omega$. Also, for every $\omega \in \Omega$, the finite triangle inequality in $\mathbb{R}$ gives
\begin{align*}
|B(\omega)|
=
\left|\sum_{i=1}^n X_i(\omega)\,\mathbb{1}_{\{|X_i|>L\}}(\omega)\right|
\leq
\sum_{i=1}^n |X_i(\omega)|\,\mathbb{1}_{\{|X_i|>L\}}(\omega)
=
C(\omega).
\end{align*}
[guided]
The purpose of the truncation is to separate the part of each $X_i$ whose magnitude is at most $L$ from the part whose magnitude is larger than $L$. For each $i \in \{1,\dots,n\}$, define the map $Y_i:\Omega \to \mathbb{R}$ by
\begin{align*}
Y_i(\omega)=X_i(\omega)\,\mathbb{1}_{\{|X_i|>L\}}(\omega).
\end{align*}
This is the large-value tail of $X_i$. The two events $\{|X_i|\leq L\}$ and $\{|X_i|>L\}$ partition $\Omega$, so their indicators satisfy
\begin{align*}
\mathbb{1}_{\{|X_i|\leq L\}}(\omega)+\mathbb{1}_{\{|X_i|>L\}}(\omega)=1
\end{align*}
for every $\omega \in \Omega$. Multiplying by $X_i(\omega)$ gives
\begin{align*}
X_i(\omega)
=
X_i(\omega)\,\mathbb{1}_{\{|X_i|\leq L\}}(\omega)
+
X_i(\omega)\,\mathbb{1}_{\{|X_i|>L\}}(\omega)
=
X_{i,L}(\omega)+Y_i(\omega).
\end{align*}
Now define $S,A,B,C:\Omega \to \mathbb{R}$ as follows. Let
\begin{align*}
S(\omega)=\sum_{i=1}^n X_i(\omega).
\end{align*}
Let
\begin{align*}
A(\omega)=\sum_{i=1}^n X_{i,L}(\omega).
\end{align*}
Let
\begin{align*}
B(\omega)=\sum_{i=1}^n Y_i(\omega).
\end{align*}
Let
\begin{align*}
C(\omega)=\sum_{i=1}^n |X_i(\omega)|\,\mathbb{1}_{\{|X_i|>L\}}(\omega).
\end{align*}
Here $S$ is the original sum, $A$ is the truncated sum, $B$ is the signed tail sum, and $C$ is the non-negative tail magnitude. Summing the identity $X_i=X_{i,L}+Y_i$ over $i$ gives the pointwise decomposition
\begin{align*}
S(\omega)=A(\omega)+B(\omega)
\end{align*}
for every $\omega \in \Omega$.
The theorem estimates the signed tail sum by the easier non-negative quantity $C$. For every $\omega \in \Omega$, applying the finite triangle inequality to the [real numbers](/page/Real%20Numbers) $Y_1(\omega),\dots,Y_n(\omega)$ gives
\begin{align*}
|B(\omega)|
=
\left|\sum_{i=1}^n X_i(\omega)\,\mathbb{1}_{\{|X_i|>L\}}(\omega)\right|
\leq
\sum_{i=1}^n |X_i(\omega)|\,\mathbb{1}_{\{|X_i|>L\}}(\omega)
=
C(\omega).
\end{align*}
This pointwise bound is the reason the final probability involves $C$ rather than $|B|$.
[/guided]
[/step]
[step:Convert the pointwise decomposition into an event inclusion]
Define the events $E,F,G\in\mathcal{F}$ as follows. Let
\begin{align*}
E=\left\{\omega \in \Omega:\ |S(\omega)|\geq t\right\}.
\end{align*}
Let
\begin{align*}
F=\left\{\omega \in \Omega:\ |A(\omega)|\geq \frac{t}{2}\right\}.
\end{align*}
Let
\begin{align*}
G=\left\{\omega \in \Omega:\ C(\omega)\geq \frac{t}{2}\right\}.
\end{align*}
We claim that $E\subset F\cup G$. Let $\omega \in E$. If $\omega \notin F$, then $|A(\omega)|<t/2$. If also $\omega \notin G$, then $C(\omega)<t/2$, and the pointwise estimate $|B(\omega)|\leq C(\omega)$ gives $|B(\omega)|<t/2$. Since $S=A+B$, the triangle inequality gives
\begin{align*}
|S(\omega)|
\leq |A(\omega)|+|B(\omega)|
<\frac{t}{2}+\frac{t}{2}
=t,
\end{align*}
contradicting $\omega \in E$. Therefore every $\omega \in E$ belongs to $F\cup G$, so $E\subset F\cup G$.
[/step]
[step:Take probabilities and use finite subadditivity]
By monotonicity of $\mathbb{P}$ and the inclusion $E\subset F\cup G$,
\begin{align*}
\mathbb{P}(E)\leq \mathbb{P}(F\cup G).
\end{align*}
By finite subadditivity of the probability measure $\mathbb{P}$,
\begin{align*}
\mathbb{P}(F\cup G)\leq \mathbb{P}(F)+\mathbb{P}(G).
\end{align*}
Substituting the definitions of $E$, $F$, and $G$ gives
\begin{align*}
\mathbb{P}\left(\left|\sum_{i=1}^n X_i\right| \geq t\right)\leq \mathbb{P}\left(\left|\sum_{i=1}^n X_{i,L}\right| \geq \frac{t}{2}\right)+\mathbb{P}\left(\sum_{i=1}^n |X_i|\,\mathbb{1}_{\{|X_i|>L\}} \geq \frac{t}{2}\right).
\end{align*}
This is the desired truncation decomposition inequality.
[/step]
