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1) Let $X_n$ and $X$ r.v. such that $X_n\to X$ in distribution (but I don't think it's important for my question). Why $$|\mathbb P\{X_n\leq x\}-\mathbb P\{X\leq x\}|\leq \mathbb P\{X_n\leq x,X>x\}+\mathbb P\{X_n>x,X\leq x\}\ \ ?$$

2) In the same spirit : Let $X_n\to X$ in distribution and $Y_n\to c$ in distribution (but still, I don't think it's important). Why, $$\mathbb P\{X_n+Y_n\leq x\}=\mathbb P\{X_n+Y_n\leq x,|Y_n-c|\leq \varepsilon\}+P\{X_n+Y_n\leq x,|Y_n-c|> \varepsilon\}\ \ ?$$

Maybe $\mathbb P\{X_n\leq x\}=\mathbb P\{X_n\leq x,X\in \mathbb R\} ?$ but since the r.v. are not independant, I have doubt for this.

and why is this $${\color{red}\leq} \ \ \mathbb P\{X_n+x\leq x+\varepsilon\}+\mathbb P\{|Y_n-c|>\varepsilon\}\ \ ?$$

Can someone explain me precisely how it work ? I alway problem with such things.

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    Because for any events $A$ and $B$, $$|P(A)-P(B)|=|P(A\cap B^c)-P(B\cap A^c)|\leqslant\max\{P(A\cap B^c),P(B\cap A^c)\}\leqslant P(A\cap B^c)+P(B\cap A^c)$$ – 2017-01-02
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    @Did: Thank you for your answer. There is something I don't understand... For you $\mathbb P\{A\cap B\}=\mathbb P\{A,B\}$ ? I really don't understand why (for example in the **2)** we start from $\mathbb P(A)$ and get $$\mathbb P\{A\}\leq \mathbb P\{A,B^c\}+\mathbb P\{A,B\}.$$ I'm not really sure that my question is clear... tell me if it is not. – 2017-01-02
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    Yes, $P(A,B)$ stands for $P(A\cap B)$. – 2017-01-02
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    thanks a lot, it's perfectly clear :) @Did – 2017-01-02
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    @Did: In an other problem, I have in a solution that $$\mathbb P\{\frac{S_n}{\sqrt n}\leq (\sigma +\delta)x,E_n\}\leq \mathbb P\{\frac{S_n}{\sqrt n}\leq x\}-\mathbb P\{E_n^c\},$$ do you agree it can't hold ? Otherwise $\mathbb P(A\cap B)+\mathbb P(B^c)\leq \mathbb P(A)$ which look strange, no ? Indeed, there is no reason to have $\mathbb P((A\cap B)\cup B^c)\leq \mathbb P(A)$ since we don't have $(A\cap B)\cup B^c\subset A$. For your information, $S_n=X_1+...+X_n$ where $X_i$ are iid, $\sigma =\sqrt{\mathbb E X_1^2}$, $E_n=\{|\frac{\sigma _n}{\sqrt n}-\sigma |\leq \delta\}$ and – 2017-01-02
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    and $\sigma _n=\sqrt{\sum_{i=1}^nX_i^2}$. But I don't think that it will help. Thank you for your consideration. – 2017-01-02

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