Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose I have a continuous real random variable $X$ and an indicator function $f(X) = \textbf{1}_{\{a \le X \le b\}}$ where $c = (a,b) \in \mathcal{A} = \{(x,y) \in \mathbb{R}^2: x <y\}$. Let $c_n=(a_n,b_n)$ where $c_n \rightarrow c$ and $f_n(X) = \textbf{1}_{\{a_n\le X \le b_n\}}$. Is it true that $f_n \rightarrow f$ almost surely?

I think it does, except perhaps on the probability zero sets $\{\omega: X(\omega) = a\}$ or $\{\omega: X(\omega)=b\}$. (Since $c_n$ goes to $c$ in two dimensions, we can't say if $a_n$ is increasing up or down to $a$, same with $b_n$)


share|cite|improve this question
Maybe it would help you think about it if you wrote $f_n(X) = 1_{(-\infty, b_n]}(X) - 1_{(-\infty,a_n)}(X)$. – cardinal Jan 6 '13 at 4:26
Yep, that makes it clear! Thanks for the suggestion. – eulerup Jan 6 '13 at 6:45
up vote 1 down vote accepted

Let $C=\mathbb{R} \setminus \{a,b \}$. Then if $x \in C$, $1_{[a_n,b_n]}(x) \to 1_{[a,b]}(x)$. So, assuming that $\mu ( X^{-1}\{a,b \} ) = 0$, then $1_{[a_n,b_n]}(X(\omega)) \to 1_{[a,b]}(X(\omega))$ a.e. $\omega$ $[\mu]$. Or, in the question's notation, $f_n \to f$ a.s.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.