why $\lim \inf \int f_n = 1?$.

Question

Im posting this problem due to inactive user

taken from here

Here is the outlines

let $f_n = \chi_{[n,n+1]}$. Then $\lim \inf f_n = 0$, and $\lim \inf \int f_n = 1$.

My question is that why $\lim \inf \int f_n = 1?$.

My attempt :Here $$f_n =\begin{cases} 1 \ \text{if }\ x\in [n,n+1] \\ 0 \ \text{if} \ x \notin [n, n+1] \end{cases}$$

I can easily see that inf $f_n=0$ and sup $f_n=1$

Im not getting why $\lim \inf \int f_n = 1?$.

What is the integral of the characteristic function of a set? — Kavi Rama Murthy, Jan 12 '21 at 09:59
For all $n$ the integral of $f_n$ is equal to $1$. Hence $$\liminf (\int f_n) = \liminf (1) = 1$$ — Crostul, Jan 12 '21 at 10:01
okk @Kavi sir u mean $$\int \chi_{[n,n+1]} d\mu = \mu([n,n+1])=n+1-n=1?$$ — jasmine, Jan 12 '21 at 10:06

score 1 · Accepted Answer · answered Jan 12 '21 at 10:03

1

Hint:

Try calculating $\int f_n$ for a couple of values of $n$...

answered Jan 12 '21 at 10:03

5xum

1 Answers1