$X_n+\tau Y\to_D X+\tau Y$ for every fixed positive constant $\tau$ implies $X_n \to_D X$. Let $X, Y, X_n's$ be random variables for which $X_n+\tau Y\to_D X+\tau Y$ for every fixed positive constant $\tau$. Show that $X_n \to_D X$.
I dont think we can let $\tau\to0$ and claim the result. Any hint on this question?
 A: Write
$$\Bbb E[e^{itX_n}] - \Bbb E[e^{itX}] =   \bigg(\Bbb E[e^{it X_{n}}] - \Bbb E[e^{it X_{n} + i t \epsilon Y}]\bigg) + \bigg(\Bbb E[e^{it X_{n} + i t \epsilon Y}] - \Bbb E[e^{it X + i t \epsilon Y}]\bigg)+ \bigg(\Bbb E[e^{it X + i t \epsilon Y}]-\Bbb E[e^{it X}]\bigg).$$
Now let $n \to \infty$ in the right hand side. By the assumption of the problem statement, the middle term goes to zero. Note that $|e^{itX_n + i t\epsilon Y} - e^{i t X_n} | = |e^{i t\epsilon Y}-1| \leq \min\{2,t\epsilon|Y|\},$ and the same bound remains true if we replace $X_n$ by $X$. Thus the first and third terms are bounded in absolute value by $\Bbb E[\min\{2,\epsilon t |Y|\}]$. Hence we showed that
$$\limsup_{n \to \infty} \big|\Bbb E[e^{i t X_n}] - \Bbb E[e^{it X}]\big| \leq 2 \Bbb E[ \min \{2,\epsilon t |Y|\}].$$
Here $\epsilon$ is arbitrary and it only appears on the right hand side, so we can let it tend to zero, and by DCT the right side converges to zero while the left side remains unchanged.
A: This is just to complement the answer by Shalop and  to address a comment posted by the OP.
The first and third are bounded by a small term:
$$
\begin{align}
\Big|\Bbb E[e^{it X_{n}}]-\Bbb E[e^{it X_{n} + i t \epsilon Y}]\Big|&\leq \Bbb E\big[\big|e^{itX_n}\big(1-e^{it\varepsilon Y}\big)\big|\big]\leq\mathbb{E}\big[|1-e^{it\varepsilon Y}|\big]\leq \Bbb E[\min(2,\varepsilon t|Y|)]\\
\Big|\Bbb E[e^{it X}]-\Bbb E[e^{it X + i t \epsilon Y}]\Big|&\leq \Bbb E\big[\big|e^{itX}\big(1-e^{it\varepsilon Y}\big)\big|\big]\leq\mathbb{E}\big[|1-e^{it\varepsilon Y}|\big]\leq \Bbb E[\min(2,\varepsilon t|Y|)]
\end{align}
$$
Here, we use the fact that for any $n\in\mathbb{Z}_+$ and $x\in\mathbb{R}$
$$
\Big|e^{ix}-\sum^n_{k=0}\frac{(ix)^k}{k!}\Big|\leq \min\Big(
\frac{|x|^{n+1}}{(n+1)!},\frac{2|x|^n}{n!}\Big)\\
$$
Dominated convergence implies that $\lim_{\varepsilon\rightarrow0}\Bbb E[\min(2,\varepsilon t|Y|)]=0$
