Clarification on convergence in distribution $\implies$ convergence in probability requiring a constant probability space

Question

Let $X_n,n\geq1$ be a sequence of random variables defined on the same probability space. Show that if $X_n \to c$ in distribution, where $c$ is a constant, then also $X_n \to c$ in probability.

I am attempting to formulate an efficient proof for this claim, which I have done as follows:

Letting $F_n$ denote the distribution function for $X_n$, and $F$ is that of $c$, i.e. $$F(x)=\begin{cases} 0\;\;x\lt c\\ 1\;\;x\geq c \end{cases}$$ we have that $F_n(x)\to F(x)$ for all $x\in\mathbb{R}\setminus\{c\}$. Choose $\epsilon\gt 0$. Then $$\begin{align}\mathbb{P}(\lvert X_n-c\rvert\lt\epsilon)=\mathbb{P}(\{X_n\lt c+\epsilon\}\setminus\{X_n\leq c-\epsilon\})\\ =\mathbb{P}(X_n\lt c+\epsilon)-\mathbb{P}(X_n\leq c-\epsilon)\\ =F_n(c+\epsilon)-F_n(c-\epsilon)\\ \to1-0=1. \end{align}$$ Note $\epsilon$ arbitrary so the result follows.

My question is what is what is the precise reason / justification for the variables to be on the same probability space? At which step would this need to be used?

Answer 1

For “$\mathbb P(\lvert X_n-c\rvert<\epsilon)$” (and other) to even make sense, the sets $$\{\lvert X_n-c\rvert<\epsilon\}=\Bigl\{\omega\in\Omega : c-\epsilon< X_n(\omega)<c+\epsilon\Bigr\}=X_n^{-1}\bigl\langle\bigl(c-\epsilon,c+\epsilon\bigr)\bigr\rangle$$ must be events of the underlying $\sigma$-algebra $\mathcal F$ on which the probability measure $\mathbb P$ is defined. One way to guarantee this is that each $X_n\colon(\Omega,\mathcal F)\to\mathbb R,\,n\in\mathbb N,$ is measurable (i.e., a random variable defined on $\Omega$).

Answer 2

Look at the definition of convergence of probability. This only makes sense if all random variables are defined on the same probability space. While this is not apparent here, but if instead of $c$ you have a random variable $X$ you need this to be even able to write $X_n-X$.

In this situation this would work also if this wasn’t the case, though. (although in these cases you’d get a sequence of probability measures $\mathbb P_n$ in the whole thing).

Answer 3

Whatever you have written is valid even if the r.v's are defined on different spaces provided $\mathbb P$ is intepreted appropriately. But convergence in probability is defined only the r.v's they are defined on the same space. For many theorems involving convergence in probability it is important that we are working within one space. For example for the fact $X_n \to X$ in probability implies $X_{n_k} \to X$ almost surely for some subsequence ${n_k}$ it is essential that $X_n$'s are defined on the same space.