[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*}[/step]

[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: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]