Sufficient conditions for: convergence in probability implies convergence in $L^p$? Is this specific condition sufficient for the mentioned statement,
Convergence in probability  $X_n \rightarrow X$ implies convergence in mean $L^p$ when the probability density function (PDF) of $X_n$ be such that: 
$$f_{X_n}(x) = 0, ~\text{for} \quad |x| > x_0  ~ \forall ~n>N$$
for some $N > 0$.
 A: W.l.o.g., assume that $N=0$. First, Denote $Y_n =|X_n-X|$, since $X_n\to X$ in probability we have $Y_n \to 0$ in probability. We know that the congergence in probability implies the convergence in distribution, that is $X_n \to X$ in distribution and it follows $\lim_{n\to \infty} \mathbb P(X_n \in [-x_0, x_0])= \mathbb P(X\in [-x_0, x_0])$. It follows from the assumption $\mathbb P(X_n \in [-x_0, x_0])=1$ that $\mathbb P(X\in [-x_0, x_0])=1$ and hence $|X|\leqslant x_0$ a.s.
Next, for $\epsilon >0$ arbitrarily,
$$\sup_{n}\mathbb E|Y_n|^{1+\epsilon}=\sup_{n}\mathbb E |X_n-X|^{1+\epsilon} \leqslant C_{\epsilon}(\sup_{n}\mathbb E|X_n|^{1+\epsilon} + \mathbb E|X|^{1+\epsilon})$$$$\leqslant C_{\epsilon} \sup_{n}\mathbb E|X_n|^{1+\epsilon} + C_\epsilon x_0^{1+\epsilon}.$$
On the other hand, 
$$\mathbb E|X_n|^{1+\epsilon} =\int_{-x_0}^{x_0}|x|^{1+\epsilon}f_n(x)dx\leqslant x_0^{1+\epsilon}\int_{-x_0}^{x_0}f_n(x)dx = x_0^{1+\epsilon} \;\mbox{ for all } n$$
as the density function $f_n\geqslant 0$ on $\mathbb R$. Combining with the argument above we get $\sup_{n}\mathbb E |Y_n|^{1+\epsilon} <\infty$ for all $\epsilon>0$. 
Now for each $p\geqslant 1$, Using Vallée-Poussin's theorem, we get that $(|Y_n|^p)_n$ is uniformly integrable (e.g., choosing $G(t)=|t|^{p+1}$ and then $\epsilon =p^2+p-1$). 
Note again that $|Y_n|^p\to 0$ in probability then $|Y_n|^p\to 0$ in $L^1$ by Vitali's theorem, or equivalently $|Y_n|\to 0$ in $L^p$ for all $p\geqslant 1$. 
Therefore, with your settings, we get a stronger conclusion is that $X_n \to X$ in $L^p$ for all $p\geqslant 1$.
A: Answer is yes (follows directly from bounded convergence). Here's an elementary argument that only relies on the definition.  
Convergence in probability means
$$ P(|X_n -X|>\epsilon) \to 0$$.
First we show that this implies that $|X|\le x_0$ a.s. If this is clear, skip to next paragraph. Otherwise:
\begin{align*}  P(|X|>x_0+\epsilon) &\le P(|X-X_n| +|X_n|>x_0+\epsilon) \\&=P(|X-X_n|+|X_n|>x_0+\epsilon,|X_n-X|\le \epsilon) + P(|X_n-X|>\epsilon)\\
& = P(\epsilon+|X_n|>x_0+ \epsilon) + P(|X_n-X|>\epsilon)\\
& = 0 + P(|X_n-X|>\epsilon)\\& \to 0.
\end{align*} 
Next estimate: 
\begin{align*} 
E[|X_n - X|^p]&=E[|X_n-X|^p,|X_n-X|>\epsilon]+\epsilon^p P(|X_n-X|\le \epsilon).\\
& \le (2x_0)^pP(|X_n-X|>\epsilon) + \epsilon^p.
\end{align*} 
Send $n\to\infty$ to obtain  
$$\limsup_{n\to\infty} E|[X_n-X|^p]=\epsilon^p.$$
But $\epsilon$ is arbitrary, and result follows. 
