# Pointwise Convergence everywhere and almost everywhere

i'm stuying measure theory.

Studying various convergence concepts, i confused with a funciton.

And this came from the "Egorov's theorem" and from Wikipedia

https://en.wikipedia.org/wiki/Egorov%27s_theorem

Here is the question,

" the indicator function $f_n(x)$$=$ $1_{[n,n+1]}(x)$,$\qquad n\in\mathbb{N}$ $\ x\in\mathbb{R}$

converges pointwise to the zero function everywhere "

but i can't understand why this indicate function coverges everywhere not almost everywhere.

when $n$ goes to $\infty$, how the interval [n,n +1] can be defined? I can't get the clear understanding..

• Converges everywhere implies converges almost everywhere. – user223391 Nov 26 '16 at 1:20

Pointwise convergence means you fix $x$ and let $n$ go to $\infty$. For each $x\in \mathbb{R}$, there is some $N>x$. Thus $f_n(x)=0$ for all $n>N$.
It converges pointwise everywhere because this works for every real $x$. As noted, everywhere implies almost everywhere.
The interval $[n,n+1]$ could be thought of as going to $[\infty,\infty]$, which is not a subset of the reals. Essentially this sequence of intervals diverges.