# If $X_n \to_{d} X$ and $Y_n\to_p 0$ then $X_n+Y_n\to_d X.$ (Convergence in Probability and Distribution)

If $$X_n \to_{d} X$$ and $$Y_n\to_p 0$$ then $$X_n+Y_n\to_d X,$$ where $$X_n\to_d X$$ means $$X_n$$ converges to $$X$$ in distribution and $$Y_n\to_p 0$$ means that $$Y_n$$ converges to $$0$$ in probability.

In order to show this let $$Z_n= X_n+Y_n$$ then $$F_{Z_n}(t) = P(X_n+Y_n\leq t)$$ $$= P(X_n+Y_n\leq t\cap |Y_n|\leq \epsilon) + P(X_n+Y_n\leq t\cap |Y_n|> \epsilon)$$ $$\leq P(X_n\leq t+\epsilon) + P(|Y_n|>\epsilon) \to F_{X}(t+\epsilon).$$

This shows that $$\limsup F_{Z_n}(t)\leq F_X(t+\epsilon).$$ I could try something similar for finding a lower bound for $$\liminf F_{Z_n}(t)$$, though I am not sure whether I am on the right path. Any hints will be much appreciated.

You showed $$P(X_n + Y_n \le t) \le P(X_n \le t + \epsilon) + P(|Y_n| > \epsilon).$$ A similar argument yields $$P(X_n \le t - \epsilon) \le P(X_n + Y_n \le t) + P(|Y_n| > \epsilon).$$
The first inequality implies $$\limsup_{n \to \infty} F_{Z_n}(t) \le F_X(t+\epsilon)$$ for any $$\epsilon > 0$$, while the second implies $$\liminf_{n \to \infty} F_{Z_n}(t) \ge F_X(t-\epsilon)$$ for any $$\epsilon > 0$$.
Taking $$\epsilon \to 0$$ yields $$F_{Z_n}(t) \to F_X(t)$$ for any continuity point $$t$$ of $$F_X$$; this is precisely the definition of convergence in distribution.