# Why: $f_n = \mathbb{1}_{(n,n+1)} \to 0$ pointwise?

I am reading the book of Folland, Real analysis, and in many cases I see as a fact that, $f_n = \mathbb{1}_{(n,n+1)} \to 0$ pointwise ($\mathbb{1}$ denotes the indicator function). I simply do not understand why this sequence converge to zero. Can anyone explain this? Thanks!

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Let $\epsilon >0$,$x\in \Bbb R$ be given. Can you find $N$ such that whenever $n>N$, $|1_{(n,n+1)}(x)|<\epsilon$? – Pedro Tamaroff Jan 11 '13 at 23:37

When you look at the (sequence of) functions as a "whole", you see that the graph of $f_n$ always has the shape of a 'square' located somewhere. This is the source of your intuition that $f_n$ shouldn't converge to zero.

However, pointwise convergence only qualifies what happens locally. Actually, it's scope is even narrower than that: it only describes what happens to a single point. And if you pick any point $\xi$ and look at $f_n(\xi)$, you'll see it is always zero, or it is briefly $1$ before becoming always 0. Thus, $f_n$ it converges pointwise to zero.

In particular, this means your intuition about the behavior of $f_n$ does not correspond well to pointwise convergence. In order to capture your intuitive notion, you'll need a different notion of convergence (e.g. uniform convergence). And in order to understand pointwise convergence, you'll need to build up new intuition regarding it.

Sequences that slide the "interesting" part off to infinity without diminishing it can actually be pretty useful. Remember this 'trick'.

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For a fixed $x$, what happens when $n > x$?

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