Convergence of indicator function of an interval

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 $b_n$)

Thanks!

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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 at 4:26
Yep, that makes it clear! Thanks for the suggestion. –  eulerup Jan 6 at 6:45
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.