Below is an example from Billingsley's Convergence of Probability Measures. $F$ here is the distribution function of the possible weak limit. I have difficulty understanding the bolded statement. For odd $P_n$, each distribution would grow linearly to $1/3$ from $-n$ to $0$ then take a jump of $1/3$ at $0$ then grow linearly to $1$. In this case, I cannot see why the possible limit should be $1/3$ for negative $x$ and $2/3$ for nonnegative $x$. And why should the existence of a weak limit imply $Q(-k,k) \le 1/3$? We should have $Q(-k,k) \le \lim \inf_m P_{n_i(m)}(-k,k)$ but the right side is always greater than $1/3$ for all $k$. I would greatly appreciate any help.
Let $\mu_n$ be the uniform distribution over $[-n,n]$, and take $P_n = \delta_0$ for even $n$ and $P_n = \frac{1}{3}\delta_0 + \frac{2}{3} \mu_n$ for odd $n$. Then $\{P_{n_i}\}$ contains a weakly convergent subsequence if $n_i$ is even for infinitely manyh $i$, but what if $n_i$ runs through odd integers only? Then the single possible limit $F(x)$ is $\frac{1}{3}$ for $x<0$ and $\frac{2}{3}$ for $x \ge 0$, and again $P_{n_i(m)} \Rightarrow_m Q$ is impossible: It would imply $Q(-k,k) \le \frac{1}{3}$ for all $k$.
Let $S_x=(-\infty,x]$. For fixed $x$ one can check that $\lim_{n\to\infty} \mu_n(S_x)=1/2$ because if $n>|x|$, we have $\mu_n(S_x)=(n-x)/(2n)$. Clearly $\delta_0(S_x)=0$ if $x<0$ and $\delta_0(S_x)=1$ if $x>0$. Since $P_n=(1/3)\delta_0+(2/3)\mu_n)$, we have $\lim P_n(S_x)= (1/3)0+(2/3)(1/2) = 1/3$ or $=(1/3)1+(2/3)(1/2)=2/3$, depending on the sign of $x$.
The villain here is $\mu_n$, which has "mass escaping to infinity", which is the opposite of what tightness and weak convergence is about.