0
$\begingroup$

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..

$\endgroup$
1
  • $\begingroup$ Converges everywhere implies converges almost everywhere. $\endgroup$
    – user223391
    Nov 26, 2016 at 1:20

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ thanks. i understood perfectly with your answer! $\endgroup$
    – user393127
    Nov 26, 2016 at 4:28

You must log in to answer this question.