Prove if {$X_i$} to converge locally in measure to $X$ and $X_i$ are independent then $X$ is constant.

Question

Prove if $\forall \varepsilon > 0 : P(\{\omega \in \Omega : |X_i(\omega)-X(\omega)| > \varepsilon \}) \underset{n \to \infty}{\to} 0$ and $X_i$ are independent then $X$ is constant.

Answer 1

Note that since $X_i \to X$ in measure, then there exists a subsequence $$(X_{i_n})_{n\in\mathbb N} \subset (X_{i})_{i\in\mathbb N},$$ such that $X_{i_n} \to X$ a.e. (see Convergence in measure implies convergence almost everywhere of a subsequence).

Therefore, we can assume without loss of generality that $X_i$ are independent and $X_i \to X$ $\mathbb P$-a.e..

Let $I = (a,b)$ be an open and bounded interval of $\mathbb N.$ It is clear (from the convergence almost everywhere) that for every $\varepsilon >0,$ $$X^{-1}(a,b) \supset \bigcup_{n=1}^{\infty} \bigcap_{i=n}^{\infty} X^{-1}_i(a+\varepsilon, b-\varepsilon)\ \mathbb P\text{-a.e. } (*). $$

Indeed, if $\omega \in \bigcup_{n=1}^{\infty} \bigcap_{i=n}^{\infty} X^{-1}_i(a+\varepsilon, b-\varepsilon),$ then there exists $i_0\in\mathbb N$ such that $$\omega \in \bigcap_{i=i_0}^{\infty} X^{-1}_i(a+\varepsilon, b-\varepsilon)\in X^i(\omega) \Leftrightarrow X^i(\omega)\in (a+\varepsilon, b-\varepsilon), \ \forall i\geq i_0. $$ Since $X_i \to X$ $\mathbb P$-a.e. the above equation shows us that $(*)$ holds.

This implies that $$X^{-1}(a,b) \supset \bigcup_{\ell = 1}^{\infty} \bigcup_{n=1}^{\infty} \bigcap_{i=n}^{\infty} X^{-1}_i\left(a+\frac{1}{\ell}, b-\frac{1}{\ell}\right)\ \mathbb P\text{-a.e. }. (**)$$

We will show that, in fact, the equality holds. $$X^{-1}(a,b) = \bigcup_{\ell = 1}^{\infty} \bigcup_{n=1}^{\infty} \bigcap_{i=n}^{\infty} X^{-1}_i\left(a+\frac{1}{\ell}, b-\frac{1}{\ell}\right)\ \mathbb P\text{-a.e. } (***). $$

Due to $(**)$ it is enough to show that $$ X^{-1}(a,b) \subset \bigcup_{\ell = 1}^{\infty} \bigcup_{n=1}^{\infty} \bigcap_{i=n}^{\infty} X^{-1}_i\left(a+\frac{1}{\ell}, b-\frac{1}{\ell}\right)\ \mathbb P\text{-a.e. }. $$

Let $\omega \in \Omega$ such that $\omega \in X^{-1}(a,b)$ and $\lim X_i(\omega) = X(\omega).$ Then there there exists $m_0\in \mathbb N$ such that $$ X(\omega) \in \left(a + \frac{1}{m_0} ,b - \frac{1}{m_0}\right),$$ and since $\lim_{i\to \infty} X_i(\omega) = X(\omega),$ we have that $$\omega \in \bigcup_{n=1}^{\infty} \bigcap_{i=n}^{\infty} X^{-1}_i\left(a+\frac{1}{2m_0}, b-\frac{1}{2m_0}\right), $$ which implies \begin{align*} X^{-1}(a,b) &\subset \bigcup_{\ell = 1}^{\infty} \bigcup_{n=1}^{\infty} \bigcap_{i=n}^{\infty} X^{-1}_i\left(a+\frac{1}{\ell}, b-\frac{1}{\ell}\right)\ \mathbb P\text{-a.e. }, \end{align*} proving (***).

The observe that $(***)$ implies that $X$ is $$ \bigcap_{i=0}^{\infty} \sigma \left(X_i,X_{i+1}, X_{i+2}, \ldots\right)\text{-measurable}.$$ Which by the Komolgorov $0-1$ law implies that $X$ is constant.