Why does convergence in distribution imply tightness A common proof for this proposition is the following: Fix $\epsilon>0$ and $M_o>0$. There exists an N such that $P(|X_n|\geq M_o) \leq \epsilon$ for all $n\geq N$. For the rest of the $X_n$ where $n<N$ (finitely many) we find $M_n$ such that $$P(|X_n|\geq M_n) \leq \epsilon$$
We then take the maximum of all the $M_n$ $n=0,1,2,..N-1$ and we get the constant for tightness or bounded in probability.
My confusion is with proving the second statement of the proof. I have seen some cite Portmanteau lemma as justification and others mention points of continuity of the cdf.
Thank you.
 A: If I understand correctly, you are wondering why $X_{n}$ converging implies that for all $\epsilon > 0$ and $M_{0} > 0$, there exists $N$ such that $P(|X_{n}| \geq M_{0}) \leq \epsilon$ for all $n \geq N$.
Well, first of all, this is actually not true. Consider $X_{n} = 2$ and $M_{0} = 1$. Then no matter how large we select $N$, we will not get $P(|X_{n}| \geq M_{0}) < \epsilon$. What I think you meant is instead that, for all $\epsilon > 0$, there exists $M_{0} > 0$ and $N$ such that
$$P(|X_{n}| \geq M_{0}) \leq \epsilon$$ for all $n \geq N$.
To find this $M_{0}$, we consider $X$, the variable that $X_{n}$ converges to in distribution. Then, since $X$ is necessarily tight, for all $\delta > 0$ there exists $M_{0} > 0$ with
$$ P(|X| \geq M_{0}) \leq \delta. $$
Take $\delta = \epsilon / 2$ and fix this $M_{0}$.
Since $X_{n} \to X$, we can find $N$ such that
\begin{equation}
| P(|X_{n}| \geq M_{0}) - P(|X| \geq M_{0}) | \leq \delta
\end{equation}
for all $n \geq N$ (this is apparently justified by the Portmanteau Lemma, although it seems like it would follow by definition). As such, by the triangle inequality,
$$ P(|X_{n}| \geq M_{0}) \leq P(|X| \geq M_{0}) + | P(|X_{n}| \geq M_{0}) - P(|X| \geq M_{0}) | \leq 2 \delta = \epsilon $$
for all $n \geq N$.
To be clear, I've adapted this from the proof in van der Vaart's Asymptotic Statistics (chapter 2.1, theorem 2.4 - Prohonov's theorem).
