Why is this sequence of random variables pairwise independent?

Question

I have a sequence $(X_n: \Omega \to \mathbb{R})_{n=1}^\infty$ of pairwise independent random variables.

Define for $n \geq 1: X_n' := X_n I_{\{X_n \leq n\}}$ where $I_A$ is the indicator function on $A$. Is it true that $(X_n')_{n=1}^\infty$ is a sequence of pairwise independent random variables?

Intuitively, this seems true. I know that pairwise independence is preserved under a Borel transformation $g: \mathbb{R} \to \mathbb{R}$ so I tried to write

$$X_n' = g \circ X_n$$

for some suitable $g$ but did not come up with anything useful. Any hints?

Answer 1

Try $$ g_n\colon x\mapsto x\mathbf 1_{\left[-\infty,n\right]}(x). $$

Answer 2

It is immediate that $X\mathbf1_{X_n\leq n}$ is measurable wrt $\sigma(X_n)$ because it is the product of two random variables that are both measurable wrt $\sigma(X_n)$.

That is enough to conclude that also the $X_n'$ are pairwise disjoint.

There is indeed a measurable function $g:\mathbb R\to\mathbb R$ such that $X_n'=g\circ X_n$.

Let $h:\mathbb R^2\to\mathbb R$ denote the function prescribed by $(x,y)\mapsto xy$ and let $k_n:\mathbb R\to\mathbb R^2$ denote the function that is prescribed by $x\mapsto(x,\mathbf1_{(-\infty,n]}(x))$.

Both functions can be shown to be measurable so that also their composition is measurable.

Then function $g=h\circ k_n$ will do the job.