I hope my question is not too stupid.

By definition a sequence $(X_n)$ of random variables converges in probability towards the random variable $X$ if for all $\epsilon >0$ we have $$ P(\lvert X_n-X\rvert >\epsilon)\to 0\text{ as }n\to\infty. $$

My question: Why do we know that the set $\left\{\lvert X_n-X\rvert > \epsilon\right\}$ is an event at all, i.e. that we can give it a probability?


1 Answer 1


Since $X_n$ and $X$ are both measurable (they are random variables after all) and the function $f(x,y) = \lvert x-y \rvert$ preserves measurability (due to its continuity), $\lvert X_n - X \rvert$ is measurable as well.

Maybe it is better for your understanding if I argue in the following way. If $X_n$ and $X$ are measurable, then so is $X_n - X$. This is a standard result in measure theory. Then, $\{ \lvert X_n - X \rvert > \varepsilon \} = \{ X_n - X > \varepsilon \} \cup \{ X_n - X < -\varepsilon \}$. Both elements on the right hand side lie in the $\sigma$-algebra defined as part of the probability space in which $X_n$ and $X$ live. Let's call this $\sigma$-algebra $\mathcal{F}$. You know from the definition of a $\sigma$-algebra that the countable union keeps you inside the $\sigma$-algebra. So then the left hand side must be in $\mathcal{F}$, i.e. it is an event.


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