$X_n \stackrel{d}{\to}X$, $Y_n \stackrel{d}{\to} c \implies X_n+Y_n \stackrel{d}{\to} X+c$ Let $X_n\Rightarrow X$ and $Y_n\Rightarrow c$. Show that $X_n+Y_n\Rightarrow X+c$.
Prove:
There exists sequences of random variables $(X^{(*)}_n)$ and $(Y^{(*)}_n)$ such that 
$(X^{(*)}_n)$ and $X_n$ have the same distribution and $X_n\rightarrow X_{\infty}$ a.s.
$(Y^{(*)}_n)$ and $Y_n$ have the same distribution and $Y_n\rightarrow Y_{\infty}=c$ a.s.
Hence we can conclude that $X_n+Y_n\rightarrow X+c$ almost surely, which implies convergence in distribution.
Can someone take a look at it?
 A: Your proof is not correct: The random variables $(X_n^{(\ast)},Y_n^{(\ast)})$ do not have the same distribution as $(X_n,Y_n)$ and therefore $$X_n^{(\ast)}+ Y_n^{(\ast)} \stackrel{d}{\to} X+c$$ does not imply $$X_n+Y_n \stackrel{d}{\to} X+c.$$ (Note: If $X \sim X'$ and $Y \sim Y'$, then $X' + Y' \sim X+Y$ does in general not hold true.)
If you want to use this kind of argumentation, then you have to show first that $(X_n,Y_n)$ converges weakly to $(X,c)$ and then you find $(X_n^{(\ast)},Y_n^{(\ast)})$ such that $(X_n^{(\ast)},Y_n^{(\ast)}) \sim (X_n,Y_n)$ and $(X_n^{(\ast)},Y_n^{(\ast)}) \to (X,c)$ almost surely.

Alternative argumentation: By Lévy's continuity theorem, it suffices to show that
$$\mathbb{E} \exp \left( \imath \xi (X_n+Y_n) \right) \to \mathbb{E} \exp\left( \imath \, \xi (X+c) \right) \qquad \text{for all} \, \, \xi \in \mathbb{R}.$$
To this end, we write
$$\begin{align*} \left| \mathbb{E} \exp \left( \imath \xi (X_n+Y_n) \right)  - \mathbb{E} \exp\left( \imath \, \xi (X+c) \right) \right| &\leq \left| \mathbb{E} \left[ \exp(\imath \, \xi (X_n+Y_n))-\exp(\imath \, \xi (X_n+c)) \right] \right| \\ &\quad + \left| \mathbb{E} \left[ \exp(\imath \, \xi (X_n+c)) - \exp(\imath \, \xi (X+c)) \right] \right| \\ &= I_1+I_2. \end{align*}$$
We estimate the terms separately. It follows from the weak convergence $X_n \to X$ that $I_2 \to 0$ as $n \to \infty$. For the first term, we recall that $Y_n \to c$ in distribution implies $Y_n \to c$ in probability (it is crucial that $c$ is a constant!). By the triangle inequality,
$$ I_1 \leq \mathbb{E}|e^{\imath \, \xi Y_n}- e^{\imath \, \xi c}|.$$
As $|e^{\imath \, x}-e^{\imath \, y}| \leq |x-y|$ and $|e^{\imath x}| =1$ for all $x,y \in \mathbb{R}$, we obtain
$$\begin{align*} I_1 &\leq |\xi| \mathbb{E}(|Y_n-c| 1_{\{|Y_n-c| \leq \delta\}}) + 2 \mathbb{P}(|Y_n-c| \geq \delta) \\ &\leq |\xi| \delta + \mathbb{P}(|Y_n-c|>\delta). \end{align*}$$
Letting $n \to \infty$ and $\delta \to 0$ shows $I_1 \to 0$ as $n \to \infty$. This finishes the proof.
