Elaborating an example that shows that pointwise convergence is not sufficient to pass the limit under the integral sign.

Question

The example is given below:

My questions are:

1- I do not understand the definition of $f_{n}$, could anyone explain it for me, please?

2- How is $\int_{0}^{1} f_{n} = 1$?

3- What is the intuition behind this definition of $f_{n}$?

Could anyone explain this to me, please?

Answer 1

1. The book has typos as you mentioned: "$f(1/n)=n$ and $f(0)=0$..." should really be $$ f_n(1/n)=n, \quad f_n(0)=0. $$

   (The authors only say later that "Define $f\equiv 0$ on $[0,1]$".)

   The graph of the function $f_k$ is as follows:

   

2. $\int_0^1f_n(x)\,dx=1$ follows from the graph above: just consider the area of the triangle.

3. This is one of the example showing that "pointwise convergence alone is not sufficient to justify passage of the limit under the integral sign". This is one type of "escape to infinity".

   If one does not require continuity, a common used example is $f_n=n1_{[\frac{1}{n},\frac{2}{n}]}$.

Answer 2

Hints:

$1).\ $ For each integer $n$, draw the graph of $f_n$. They will be triangles of decreasing base and increasing height.

$2).\ $ Although the dimensions of the triangles are changing, the area is not.

$3).\ $ For each $x\in [0,1]$, all the $f_n's$ are eventually zero at $x$ so $\underset{n\to\infty}\lim f_n(x)=0$ but the areas are steady at a value $1$ so.....