Vtyjkyuk

Example
For each natural number $n$, define $f_n$ on $[0,1]$ to have value $0$ if $x geq 2/n$, have $f(1/n) = n$, $f(0) = 0$ and be linear on the intervals $[0,1/n]$ and $[1/n,2/n]$.
Observe that $int_0^1 f_n = 1$ for each $n$.
Define $f equiv 0$ on $[0,1]$.
Then
$$
text$f_n to f$ pointwise on $[0,1]$,
but $lim_n to infty int_0^1 f_n neq int_0^1 f$.
$$
Thus, pointwise convergence alone is not suffcient to justifty passage of the limit under the integral sign.

(Original image here.)

I am not sure that I understand this example correct. First, shouldn't $f(1/n)=n$ and $f(0) = 0$ be $f_n(1/n) = n $ and $f_n(0)$?

Also, when $n to infty$, $f_n to infty$ on $[1/n, 2/n]$, but the interval also goes to $0$. How can we deal with this situation? Should we say $f_n to 0$ on $[1/n, 2/n]$ since the interval goes to $0$?

Any help is appreciated.

You are right. It should read $f_n(frac 1 n)=n$ and $f_n(0)=0$. The precise deinition is $f_n(x)=n^2x$ for $0leq x leq frac 1 n $, $f_n(x)=n^2(frac 2 n - x)$ for $frac 1 n leq x leq frac 2 n $ and $0$ for $x geq frac 2 n$. To see that $f_n(x) to 0$ for every $x$ note that if $x>0$ then $frac 2 n <x$ for all $n$ sufficiently large, so $f_n(x)=0$ for such $n$. To show that $int f_n$ does not converge to $int 0 , dx=0$ you can draw a picture of the graph; the integral of $f_n$ is the area of a triangle with base $frac 2 n$ and height $n$ which is $1$.

You are right: it should be $f_nleft(frac1nright)=n$ and $f_n(0)=0$.

On the other hand, I don't understand your doubts concerning the passage to the limit. Let $xin[0,1]$:

1. If $x=0$; then $(forall ninmathbbN):f_n(x)=0$ and therefore $lim_ntoinftyf_n(x)=0$.


2. Otherwise, take $Ninmathbb N$ such that $frac2N<x$. Then $ngeqslant Nimplies f_n(x)=0$ and therefore $lim_ntoinftyf_n(x)=0$ too.