# Why is this inequality of truncated random variables correct?

Given a sequence $$X_k$$ of $$k$$ independent random variables, the book am reading says I can form 'truncated' random variables, i.e.

$$X_k(n) = X_k \cdot \textbf{1}_{\{ \omega : |X_k(\omega)| \leq n\}}$$

where $$\omega$$ means an outcome from the sample space and $$\textbf{1}$$ is an indicator function.

Define $$S_n$$ and $$\hat{S}_n$$ as:

$$S_n = X_1 + X_2 + ...+X_n$$

$$\hat{S}_n = X_1(n) + X_2(n) + ... X_n(n)$$

so this means that $$\hat{S}_n$$ is the sum of $$n$$ truncated random variables.

We also define $$m_n$$ as:

$$m_n = \mathbb{E}(X_1(n))$$

Since the variable distributions (of $$X_k$$) are the same, we have $$m_n=\mathbb{E}(X_k(n))$$ for all $$k \geq 1$$.

The book am reading says that the following inequality is 'obvious', where given $$\epsilon > 0$$, we have:

$$P\bigg(\big | \frac{S_n}{n} - m_n \big | \geq \epsilon\bigg) \leq P\bigg(\big | \frac{\hat{S}_n}{n} - m_n \big | \geq \epsilon\bigg) + P\bigg(\hat{S}_n \neq S_n\bigg)$$

But it's not at all obvious to me and I got lost. How can we have the above inequality? (i.e. we might get some large $$X_k$$ inside the $$S_n$$, such that the left term is much larger than the right term which is truncated).

The event $$[|S_n/n-m_n|\ge\epsilon]$$ is contained in the union of the two events $$[|\hat S_n/n-m_n|\ge\epsilon]$$ and $$[\hat S_n\ne S_n]$$. Think: if some outcome is not in $$[\hat S_n\ne S_n]$$, then $$\hat S_n - S_n$$, and in that case for $$|S_n/n-m_n|\ge\epsilon$$ to occur it must also happen that $$|\hat S_n/n-m_n|\ge\epsilon$$.