how to find such a sequence of measurable functions?

Question

I want to find a sequence of measurable functions $f_n(x):[0,1]\rightarrow[0,1]$,such that $\int{f_n}\rightarrow{0}$ as $n\rightarrow\infty$, but for no $x\in[0,1]$ does $f_n(x)$ converge to a limit as $n\rightarrow\infty$.

I wanted to use a translation on this sequence and shrink it at the same time, but I got stuck. It always has a lot of points converging to zero as I want to make the integral small. Can anyone help me out?

Answer 1

Let $\phi_{n,k} = \begin{cases} 1_{[{k \over n}, {k+1\over n})}, & k=0,...,n-2 \\ 1_{[{n-1 \over n}, 1]}, & k=n-1 \end{cases}$. Then $\int \phi_{n,k} = {1 \over n}$.

Then consider the sequence $\phi_{1,0}, \phi_{2,0}, \phi_{2,1}, \phi_{2,0}, \phi_{3,1}, \phi_{3,2},...$.

Also, for all $x$ there exists $k_0,k_1$ such that $\phi_{n,k_0}( x) = 0$, $\phi_{n,k_1}( x) = 1$.

Answer 2

Try $f_n(x) = ((\sin (n(x+1)) + 1)/2)^k$ for some $k > 0$, like $k = 2$ or $k = 3$ or perhaps higher. And perhaps use $n$ in the exponent instead of $k$. You should get that the integrals go to zero (needs to be checked), but it should be that there cannot be any value $x$ where you get a limit.