Does $X_n \xrightarrow{\text{in distr.}} X$ and $|X_n|\leq Y$ imply $|X|\leq Y$? We know that $$X_n \xrightarrow{\mathbb P} X \text{ and } |X_n|\leq Y \implies |X|\leq Y \text{ a.s.}$$ I was wondering if the same holds in case of convergence in distribution. So far, I've shown that $$X_n \xrightarrow{\text{in distr.}} X \text{ and } |X_n|\leq c \in \mathbb R \implies |X|\leq c \text{ a.s.}$$
So the question is if the following holds: $$X_n \xrightarrow{\text{in distr.}} X \text{ and } |X_n|\leq Y \implies |X|\leq Y \text{ a.s.}$$
 A: No--Assume that $X_n=Y$ for every $n$ with $P(Y=1)=P(Y=2)=\frac12$ and that $X=3-Y$. Then $|X_n|\leqslant Y$ almost surely, for every $n$, and the distributions of $X$ and $Y$ coincide hence $X_n\to X$ in distribution, but $[|X|\leqslant Y]=[Y=2]$ has probability $\frac12$ only.
Extending this to $k$ values instead of $2$ allows the probability of $[|X|\leqslant Y]$ to be as small as one wants but positive. Exercise: $P[|X|\leqslant Y]=0$ is impossible.
A: No, this implication does not hold. Roughly speaking: Convergence in distribution does not preserve pointwise properties of random variables.
Consider $([0,1],\mathcal{B}([0,1]))$ endowed with the Lebesgue measure and define $$\begin{align*} X(x) &:= \begin{cases} 0 & x < \frac{1}{2} \\ 1 & x \geq \frac{1}{2} \end{cases} \\ Y(x) &:= \begin{cases} 1 & x < \frac{1}{2} \\ 0 & x \geq \frac{1}{2} \end{cases} \end{align*}, \qquad x \in [0,1].$$
If we set $X_n := Y$, then $X_n \to X$ in distribution and $|X_n| \leq |Y|$, but $|X_n| \leq |X|$ does not hold.
