$\int f_k\to 0 $ but $f_k $ does not converge to $0 $ ae, where $ f_k $ is defined in $[0, 1] $

Question

Give an exemple, in [0, 1], of a sequence of functions $ f_k $ such that $||f_k||_ 1=\int |f|_k \to 0 $ but $ f_k $ does not converge to $0 $ a.e.

Answer 1

Picture: Choose $f_k$ to be a box with length "about" $1/k$, and move the box around the interval.

More rigorous: Fix an enumeration of the rationals in $[0, 1]$ by $\{0, 1, 1/2, 1/3, 2/3, 1/4, 3/4, ...\}$, and let

$$f_k = \chi_{(r_k - 1/k, r_k + 1/k) \cap [0, 1]}$$

where $r_k$ is the kth entry in the sequence. Then $f_k$ converges nowhere, but

$$\int |f_k| \leq \frac{2}{k}$$

Answer 2

Try to consider the functions

$$f_{n,k}=\begin{cases}n,&x\in \left[\frac{k}{n^2},\frac{k+1}{n^2}\right],\\0,&\text{otherwise}\end{cases}$$ with $n\in \Bbb N$, $k=0,\dots,n^2-1$.

Answer 3

Given a set $A$ recall that $\mathcal{X}_A$ is the characteristic function of $A$.

Let $f_1=\mathcal{X}_{[0, 1]}$

$f_2=\mathcal{X}_{\left[0, \frac{1}{2}\right]}$

$f_3=\mathcal{X}_{\left[\frac{1}{2}, 1\right]} $

$f_4=\mathcal{X}_{\left[0, \frac{1}{3}\right]} $

$f_5=\mathcal{X}_{\left[\frac{1}{3}, \frac{2}{3}\right]} $

$f_6=\mathcal{X}_{\left[\frac{2}{3}, 1\right]} $

$f_7=\mathcal{X}_{\left[0, \frac{1}{4}\right]} $

and so on...

Answer 4

Define $I_{n,k}:=[k2^{-n},(k+1)2^{-n})$ for $n\in\mathbb N$ and $0\leqslant k\leqslant 2^n-1$. Then define $$f_1=a_1\chi_{I_{1,0}}, f_2:=a_1\chi_{I_{1,1}}$$ $$f_3=a_2\chi_{I_{2,0}}, f_4=a_2\chi_{I_{2,1}}, f_5=a_2\chi_{I_{2,2}}, f_6=a_2\chi_{I_{2,3}}$$ $$\vdots$$ $$f_{2^{n+1}+1+k}:=a_n\chi_{I_{n,k}},\quad 0\leqslant k\leqslant 2^n-1.$$ Taking $a_n$ such that $a_n2^{-n}\to 0$, we have $\lVert f_n\rVert_{L^1}\to 0$, but there is not convergence almost everywhere to $0$ (because for each $x$, there is $A(x)\subset \mathbb N$ infinite such that $f_j(x)>1$ for all $j\in A(x)$.

However, it is true that if $f_n\to 0$ in $L^1$, then there is $n_k\uparrow \infty$ such that $f_{n_k}\to 0$ almost everywhere: we consider $n_k$ such that $\mu\{|f_{n_k}|>2^{-k}\}\lt 2^{-k}$. It holds for any measure space, not only on the unit interval endowed with Lebesgue measure.