I need an example of a sequence of integrable functions on $[0,1]$ s.t. $$\lim_{n\to\infty}\int_0^1 |f_n(x)|\,dx = 0$$
but $f_n$ does not converge to $0$ a.e.
Anyone can provide an example with a detailed explanation? thanks.
I know about the indicator function example that splits $[0,1]$ as $[0,\frac{1}{2}],[\frac{1}{2},1]$ and $[0,\frac{1}{3}]$... I want to see a different one. Thanks.
On
Take the indicator/characteristic function of $[0,1/2]$ for $f_1$, and the same over the interval $[1/2,1]$ for $f_2$, and the same over the interval $[0,1/3]$, for $f_3$, as well as the indicator over $[1/3,2/3]$ for $f_4$, etc.
In this way, clearly the integrals are converging to $0$ since the areas of the rectangles are getting smaller and smaller and smaller. But $f_n$ doesn't converge to $0$ almost everywhere (why?).
On
Let $f_1(x)=1_{[0,1]}(x)$ (which is identically $1$ on $[0,1]$).
Let $f_2(x)=1_{[0,1/2]}(x)$, and $f_3(x)=1_{[1/2,1]}(x)$.
Let $f_4(x)$, $f_5(x)$, $f_6(x)$, and $f_7(x)$ be the indicators of $[0,1/4]$, $[1/4,1/2]$, $[1/2,3/4]$, and $[3/4,1]$, respectively.
And in general: define $f_{2^n}(x)$ through $f_{2^{n+1}-1}(x)$ be the indicators of $[0,1/2^n],\ldots,[1-1/2^n,1]$, respectively.
Then $$ \int_0^1f_n(x)\,dx=\frac{1}{2^m},\qquad 2^m\leq n<2^{m+1}, $$ so that $\int_0^1 f_n(x)\,dx\to0$ as $n\to\infty$. But, every single $x\in[0,1]$ satisfies $f_n(x)=1$ for infinitely many $n$.
For $n\ge 1$ and $0\le k<2^n$ $$f_{2^n+k} \left(x\right) = \chi_{\left[\frac{k}{2^n}, \ \frac{k+1}{2^n} \right]} \left(x\right).$$
when $\chi_{\left[a,b\right]}$ is the indicator function of $\left[a,b\right]$.
And then $$\lim_{2^n+k\to\infty}\int_{\left[0,1\right]}\left|f_{2^n+k}\left(x\right)\right|dx= \lim_{2^n+k\to\infty}\frac{1}{2^n}=\lim_{n\to\infty}\frac{1}{2^n}=0$$
And also for all $x\in\left[0,1\right]$ $$\limsup_{n\to\infty}f_n\left(x\right)=1,$$ which means, in particular, that $f_n$ does not converge to $0$ almost surely.