A degenerate weak limit with persistent expectation

Set ${\mathrm{\Omega }}_n=\{0,1\}$, ${\mathcal{A}}_n=\mathcal{P}({\mathrm{\Omega }}_n)$, $P_n(\{1\})=1\mathrm{/}n$, and $P_n(\{0\})=1-1\mathrm{/}n$. These masses are nonnegative and sum to one. Since the space has two points, every disjoint sequence of subsets has at most two nonempty members; the mass formula is countably additive and gives zero to the empty set by [F3]. Thus [F1], [F2], and [F4] make this a probability space for each $n$. The map $X_n(0)=0$, $X_n(1)=n$ is measurable by [F5], hence a real random variable by [F6].

For any fixed $\epsilon >0$, once $n>\epsilon$ the event $\{\mathrm{\mid }{X}_n\mathrm{\mid }>\epsilon \}$ is exactly $\{1\}$, so $P_n(\mathrm{\mid }{X}_n\mathrm{\mid }>\epsilon )=1\mathrm{/}n\to 0$. Therefore $X_n=o_P(1)$ by [F7].

Fix bounded continuous $f:\mathbb{R}\to \mathbb{R}$, set $M=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, and put $g=f+M\ge 0$. By [F13], the law $P_{X_n}$ assigns mass $1-1\mathrm{/}n$ to $0$ and $1\mathrm{/}n$ to $n$. Its pushforward formula makes it countably additive because preimages of disjoint Borel sets are disjoint, and $P_{X_n}(\mathrm{\varnothing })=P_n(\mathrm{\varnothing })=0$; its total mass is $P_n({\mathrm{\Omega }}_n)=1$, so it is a probability measure by [F3]–[F4]. The point mass ${\delta }_0$ is also a Borel probability measure: it assigns zero to $\mathrm{\varnothing }$, and in a disjoint family of Borel sets at most one set contains $0$, while ${\delta }_0(\mathbb{R})=1$. Since $\mathrm{\mid }f\mathrm{\mid }\le M$ and $0\le g\le 2M$, every nonnegative simple minorant of $\mathrm{\mid }f\mathrm{\mid }$ or $g$ has simple integral at most the corresponding constant times $\mu (\mathbb{R})$, by finite additivity and [F12]. Taking the suprema in the nonnegative integral defining expectation [F11] gives $\int \mathrm{\mid }f\mathrm{\mid }d\mu \le M$ and $\int gd\mu \le 2M$ for either probability law $\mu \in \{P_{X_n},{\delta }_0\}$, so both functions are integrable. Thus [F15] gives $\int fd\mu =\int (g-M)d\mu =\int gd\mu -M$. The nonnegative simple function $g(X_n)$ has expectation $\int gd{P}_{X_n}=(1-1\mathrm{/}n)g(0)+(1\mathrm{/}n)g(n)$ by [F11]–[F12] and [F14]; the same formula gives $\int gd{\delta }_0=g(0)$. Thus $\mathrm{\mid }\int fd{P}_{X_n}-\int fd{\delta }_0\mathrm{\mid }=\mathrm{\mid }\mathbb{E}g(X_n)-g(0)\mathrm{\mid }=\mathrm{\mid }g(n)-g(0)\mathrm{\mid }\mathrm{/}n\le 2M\mathrm{/}n\to 0$. By [F8], the laws converge weakly to ${\delta }_0$; by [F9]–[F10], the asymptotic distribution is ${\delta }_0$.

The nonnegative simple-integral formula [F11]–[F12] gives $\mathbb{E}{X}_n=(1\mathrm{/}n)n+(1-1\mathrm{/}n)0=1$ for every $n$, while the limit variable $0$ has expectation $0$. For the finite check $n=2$, $P_2(X_2=2)=1\mathrm{/}2$ and $\mathbb{E}{X}_2=1$, confirming the stated mass and expectation formulas.