Slutsky's Theorem XnYn-CX

Question

My Question is How they got step c and Why d can conclude e step.

Answer 1

For step (c), let $E_1$ denote the event $\{x < X_n \leq y\}$, and let $E_2$ denote the event $\{\left| Y_n \right| < \varepsilon / M\}$. Then we have \begin{align*} \mathbb{P}(x < X_n \leq y, \left| Y_n \right| < \varepsilon / M) &= \mathbb{P}(E_1 \cap E_2)\\ &= \mathbb{P}(E_1) + \mathbb{P}(E_2) - \mathbb{P}(E_1 \cup E_2)\\ &= [1 - \mathbb{P}(E_1^c)] + [1 - \mathbb{P}(E_2^c)] - \mathbb{P}(E_1 \cup E_2)\\ &= 1 - \mathbb{P}(E_1^c) - \mathbb{P}(E_2^c) + [1 - \mathbb{P}(E_1 \cup E_2)]\\ &\geq 1 - \mathbb{P}(E_1^c) - \mathbb{P}(E_2^c)\\ &= 1 - \mathbb{P}(\{x < X_n \leq y\}^c) - \mathbb{P}(\{\left| Y_n \right| < \varepsilon / M\}^c)\\ &= 1 - \mathbb{P}(\{x < X_n \leq y\}^c) - \mathbb{P}(\left| Y_n \right| \geq \varepsilon / M), \end{align*} where $"\geq"$ comes from the fact that the probability of any event ($E_1 \cup E_2$ in this case) is less or equal to 1.

For step (e), given that we have concluded the result from step (d), i.e. $$\lim_{n\rightarrow \infty} \mathbb{P}(x < X_n \leq y, \left| Y_n \right| < \varepsilon / M) > 1 - \xi.$$ Together with the result from step (b) and the monotonicity of limit, we have $$1 \geq \lim_{n\rightarrow \infty}\mathbb{P}(\left| X_nY_n \right| < \varepsilon) \geq \lim_{n\rightarrow \infty}\mathbb{P}(x < X_n \leq y, \left| Y_n \right| < \varepsilon / M) > 1 - \xi.$$ Since this holds for any $\xi > 0$, using the Squeeze Theorem, we conclude that $$\lim_{n\rightarrow \infty}\mathbb{P}(\left| X_nY_n \right| < \varepsilon) = 1,$$ which then implies that $$\lim_{n\rightarrow \infty}\mathbb{P}(\left| X_nY_n \right| \geq \varepsilon) = 0.$$ Lastly, note that $\mathbb{P}(\left| X_nY_n \right| \geq \varepsilon) > \mathbb{P}(\left| X_nY_n \right| > \varepsilon)$. Using the monotonicity of limit, we have $$0 = \lim_{n\rightarrow \infty}\mathbb{P}(\left| X_nY_n \right| \geq \varepsilon) \geq \lim_{n\rightarrow \infty}\mathbb{P}(\left| X_nY_n \right| > \varepsilon) \geq 0.$$ Using the Squeeze Theorem once again, we can conclude that $$\lim_{n\rightarrow \infty}\mathbb{P}(\left| X_nY_n \right| > \varepsilon) = 0.$$

Remark: The last part of proof gives a rigorous justification for the fact that the inequality sign is strict (i.e. $">"$ instead of $"\geq"$) from the result in part (e). However, this would not affect the remaining proof for the Slutsky's Theorem.