Does almost sure convergence imply uniform boundedness

Question

Let $(X_{n})_{n},X$ be random variables and assume $P(X_{n}\to X)=1$

Until now I have always assumed that the sequence is uniformly bounded, i.e.

there exists $k > 0$ so that $P(\sup\limits_{n}\vert X_{n}-X\vert<k)=1(*)$

but in a solution, $P(X_{n}\to X)=1$ was mentioned as an additional assumption to $(*)$. Does this mean almost sure convergence does not imply uniform boundedness?

and what about $P(\sup\limits_{n}\vert X_{n}\vert<k)$? Is it implied that it is almost surely uniformly bounded from almost sure boundedness?

Answer 1

Hint: Consider $(0,1)$ with Lebesgue measure and $$X_n := n 1_{(0,1/n)}.$$

Answer 2

Let the statements $$ \tag{S1} \mbox{there exists a constant } K \mbox{ such that}\sup_n\left\lvert X_n-X\right\rvert\leq K\mbox{ a.s.} $$ $$\tag{S2} X_n\to X\mbox{ a.s.} $$ Then there exists

• a sequence $(X_n)$ satisfying (S1) but not (S2), namely, $X_n=(-1)^n$,
• a sequence $(X_n)$ satisfying (S2) but not (S1), namely, a sequence $(X_n)$ such that $P(X_n=n)=2^{-n}$ and $P(X_n=0)=1-2^{-n}$.