Why does $\mathbb{P}(\sum_{k=1}^{n}X_{k}\neq\sum_{k=1}^{n}X_{k}\mathbb{1}_{|X_{k}|\leq C})\leq\sum_{k=1}^{n}\mathbb{P}(|X_{k}|>C)$ for $X_{k}$ i.i.d? As asked in the title, I'd like to show 

If $X_{k}$ are i.i.d with symmetric distribution  $X_{k}=_{d}-X_{k}$, then we have $$\mathbb{P}\Big(\sum_{k=1}^{n}X_{k}\neq\sum_{k=1}^{n}X_{k}\mathbb{1}_{|X_{k}|\leq C}\Big)\leq\sum_{k=1}^{n}\mathbb{P}(|X_{k}|>C)$$

I understand the reason why we finally have $\mathbb{P}(|X_{k}|>C)$, since somehow we can change the probability into $\mathbb{P}(X_{k}\neq X_{k}\mathbb{1}_{|X_{k}|\leq C})$ which is exactly $\mathbb{P}(|X_{k}|>C)$.
I just don't understand how we move the sum outside the probability. That is, we firstly have $$\mathbb{P}\Big(\sum_{k=1}^{n}X_{k}\neq\sum_{k=1}^{n}X_{k}\mathbb{1}_{|X_{1}|\leq C}\Big)=\mathbb{P}\Big(\sum_{k=1}^{n}(X_{k}-X_{k}\mathbb{1}_{|X_{k}|\leq C})\neq 0\Big),$$ then it seems what I wanted prove suggests me have this:$$\mathbb{P}\Big(\sum_{k=1}^{n}(X_{k}-X_{k}\mathbb{1}_{|X_{k}|\leq C})\neq 0\Big)\leq \sum_{k=1}^{n}\mathbb{P}(X_{k}-X_{k}\mathbb{1}_{|X_{k}|\leq C})\neq 0),$$ but this is not correct, right?
The set of $\sum_{k=1}^{n}X_{k}\neq\sum_{k=1}^{n}X_{k}\mathbb{1}_{|X_{k}|\leq C}$ should be larger than the set of $$\{(X_{1}\neq X_{1}\mathbb{1}_{|X_{1}\leq C})\ \text{or} \ (X_{2}\neq X_{2}\mathbb{1}_{|X_{2}\leq C})\ \text{or}\ \cdots\ \text{or}\ (X_{n}\neq X_{n}\mathbb{1}_{|X_{n}\leq C})\},$$ right?
What am I missing here?
 A: 
The set of $\sum_{k=1}^{n}X_{k}\neq\sum_{k=1}^{n}X_{k}\mathbb{1}_{|X_{k}|\leq C}$ should be larger than the set of $$\{(X_{1}\neq X_{1}\mathbb{1}_{|X_{1}|\leq C})\ \text{or} \ (X_{2}\neq X_{2}\mathbb{1}_{|X_{2}|\leq C})\ \text{or}\ \cdots\ \text{or}\ (X_{n}\neq X_{n}\mathbb{1}_{|X_{n}|\leq C})\},$$ right?

No.  In words, if the sums are unequal, then at least one pair of corresponding summands must be unequal.  Therefore, $$\left\{\sum_{k=1}^n X_k \neq \sum_{k=1}^nX_k1_{|X_k|\leq C}\right\} \subseteq \bigcup_{k=1}^n\left\{X_k \neq X_k1_{|X_k|\leq C}\right\}$$

For some, it may be easier to understand this if we take complements (using De Morgan's laws) to see the equivalent inclusion
$$\bigcap_{k=1}^n\left\{X_k = X_k1_{|X_k|\leq C}\right\} \subseteq \left\{\sum_{k=1}^n X_k = \sum_{k=1}^nX_k1_{|X_k|\leq C}\right\}$$
where we have applied $$A \subseteq B \iff \Omega \setminus A \supseteq \Omega\setminus B$$ (i.e. taking complements reverses the direction on the set inclusion)
A: You've got a good start. You can continue with the following reasoning.
$$\mathbb{P}\Big(\sum_{k=1}^{n}X_{k}\neq\sum_{k=1}^{n}X_{k}\mathbb{1}_{|X_{1}|\leq C}\Big)=\mathbb{P}\Big(\sum_{k=1}^{n}(X_{k}-X_{k}\mathbb{1}_{|X_{k}|\leq C})\neq 0\Big),$$
$$ \le \mathbb{P}\Big(\cup_{k=1}^{n}\left((X_{k}-X_{k}\mathbb{1}_{|X_{k}|\leq C})\neq 0\right)\Big)$$
The last inequality was due to the fact that sum of $n$ terms is not zero leads to at least one of the $n$ items is not zero.
$$ \le \sum_{k=1}^{n}\mathbb{P}\Big((X_{k}-X_{k}\mathbb{1}_{|X_{k}|\leq C})\neq 0\Big) =\sum_{k=1}^{n}\mathbb{P}\Big(X_{k} \neq X_{k}\mathbb{1}_{|X_{k}|\leq C}\Big) = =\sum_{k=1}^{n}\mathbb{P}\Big(X_{k} > C\Big). $$
