I just read Van der Vaar's book Asymptotic Statistics on page 8 in the proof of the Prohorov's theorem. It seems to me that the author implicitly assumes that any r.v. is tight, or big-O-pee one.
My understanding is that it is because we assume any random variable only takes values in $\mathbb{R}$ (not $\mathbb{\overline{R}}$). So it cannot take $\infty$ or $-\infty$.
Is it correct?
This is a consequence of the cdf properties: for any r.v. $X$, $F_X(x)=\mathbb P(X\leq x)$ has limits at $\pm\infty$ $$ \lim_{x\to-\infty}F_X(x) = 0, \quad \lim_{x\to\infty}F_X(x) = 1 $$ So for any $\varepsilon$ there exists $M$ s.t. $\mathbb P(X<-M)<\varepsilon/2$ and $\mathbb P(X>M)=1-F_X(M)<\varepsilon/2$ and then $$\mathbb P(|X|>M)<\varepsilon.$$
In turn, these properties of the cumulative distribution function follow from the continuity property of the probability measure and the fact that the random variable does not take infinite values. Indeed, define the decreasing sequence of events $$ B_n=\{|X|>n\}, \quad B_{n+1}\subseteq B_n $$ Then $B=\bigcap_{n=1}^\infty B_n=\{|X|=\infty\}=\varnothing$, and continuity of measure property implies that $\mathbb P(B_n)\to \mathbb P(B)=0$. It allows for any $\varepsilon>0$ to chose $M$ such that $\mathbb P(|X|>M)<\varepsilon$.