{$\int_{[1/n,1]}f$} to converge and yet $f$ is not $L$-integrable over $[0,1]$

Question

Let $f$ be a function on $[0,1]$ and continuous on $(0,1]$.

I want to find a function $f$ s.t. {$\int_{[1/n,1]}f$} converges and yet $f$ is not $L$-integrable over $[0,1]$.

My attempts:

I've found $f(x)=(1/x)Sin(1/x)$ but I can not prove that $f$is not $L$-integrable over $[0,1]$.

Answer 1

You can piece together a continuous function $f$ on $(0,1]$ such that $f(1/n)=0$ for all $n$, and $f$ has a big spike down and a big spike up on $\left[\frac{1}{n+1},\frac{1}{n}\right]$, with $\int_{1/(n+1)}^{1/n}f(x)dx = 0$ and $\int_{1/(n+1)}^{1/n}|f(x)|dx = 1$.