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In this wildly popular post, there is a claim:

I like to remember this by example; specifically let $f_n = \chi_{[n,n+1]}$. Then $\lim \inf f_n = 0$, and $\lim \inf \int f_n = 1$.

So $f_n = \chi_{[n,n+1]}$ is a family of rectangles. I can appreciate that $\lim \inf \int f_n = 1$ since the area underneath is always $1$ regardless

How do you justify "$\lim \inf f_n = 0$", since $f_n$ maintains a height of $1$ always?

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Fraïssé
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  • The second $\liminf$ is taken pointwise. – Michael Burr Apr 23 '16 at 21:36
  • $\lim \inf f_n = 0$ means that, for all $x$, $\lim \inf f_n(x) = 0$. It is easy to see that for all $x\in \mathbb{R}$, $\lim \inf \chi_{[n,n+1]}(x) =0$. – Ramiro Apr 23 '16 at 23:16
  • @Ramiro What about where $\chi$ is supported? – Fraïssé Apr 23 '16 at 23:29
  • $\lim\inf f_n=0$ means that, for all $x$, $\lim\inf f_n(x)=0$. This means: for each FIXED $x$, $\lim\inf f_n(x)=0$. Now, for each fixed $x\in \mathbb{R}$, it is easy to see that $\lim\inf \chi_{[n,n+1]}(x)=0$. One must inspect what happens in each fixed $x$, not mattering what happens in other points. That is why it is called "pointwise (convergence)". – Ramiro Apr 24 '16 at 03:34

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The point-wise limit of $f_n$ as $n$ goes to $\infty$ is $0$. Its limit is a line on the $x$-axis everyhwere except at $\infty$ it is $1$.

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