# How to demonstrate that $1_{\text{inf}_{n\ge k} A_n} = \inf_{n\ge k} 1_{A_n}$?

I'm reading Resnick's "A probability path" and on page 10 it states the result:

$1_{\text{inf}_{n\ge k} A_n} = \inf_{n\ge k} 1_{A_n}$

Where $\text{inf}_{n\ge k} A_n := \bigcap_{n=k}^{\infty} A_n$

Now, suppose $\exists j$ $j > k$ such that $\exists\omega$ $\forall n \ge j$ $\omega \in A_n$ and $\forall n \lt j$ $\omega \notin A_n$. It is clear to me that $1_{\text{inf}_{n\ge k} A_n} = 1$.

However, for $n\ge k$, $1_{A_n}$ is a sequence of 0s followed by a sequnce of 1s. It seems to me that the infimum should be interpreted as the greatest lower bound of the set ${0,1}$. Thus I conclude that the L.H.S. is 0. I think my interpretation of the LHS is wrong. What is my error?

Note that $\inf_{n \geq k} \mathbb{1}_{A_n}(x)=1$ if and only if $x \in A_n$ for all $n\geq k$, so $x\in \bigcap_{n=k}^{\infty}A_n$. Otherwise if there happened to be a $n^*\geq k$ such that $x\not\in A_{n^*}$, $\mathbb{1}_{A_{n^*}}(x)=0$, and the infimum would be zero.
On the other hand, $\mathbb{1}_{\inf_{n\geq k}A_n}(x) = \mathbb{1}_{\bigcap_{n=k}^{\infty}A_n}(x)=1$ if and only if $x\in \bigcap_{n=k}^{\infty}A_n$. Otherwise, $x\not\in \bigcap_{n=k}^{\infty}A_n$ if and only if $\mathbb{1}_{\bigcap_{n=k}^{\infty}A_n}(x)=0$.
Hence, we see $\left(\inf_{n \geq k} \mathbb{1}_{A_n}(x)=1\right)$ if and only if $\left(x\in \bigcap_{n=k}^{\infty}A_n\right)$ if and only if $\left(\mathbb{1}_{\bigcap_{n=k}^{\infty}A_n}(x)=1\right)$, which is what you wanted.