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

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 $[0,1]$.

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Oops! I understand what you mean now by $\{\int_{[\frac1n,1]}f\}$ converges. I have deleted my answer. – Cameron Buie Nov 5 '12 at 8:03

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 $\int_{1/(n+1)}^{1/n}|f(x)|dx = 1$.
I think my example has characteristics of what you saying, but why it is not L-integrable on $[0,1]$? – Bunny Nov 5 '12 at 8:16
@Bunny: Any function $f$ satisfying the description in my answer is not Lebesgue integrable, because $\int|f|=+\infty$, or because for all $n$ there exists $a_n\in\left(\frac{1}{n+1},\frac{1}{n}\right)$ such that $\left|\int_{a_n}^1 f\right| = \frac{1}{2}$, while $\int_{1/n}^1 f = 0$. – Jonas Meyer Nov 5 '12 at 8:24
@JonasMeyer: $f$ is just some function, how can it depend on $n$? – akkkk Nov 5 '12 at 8:28
Ah, I get it, it has note one big spike but denumerably many - one for each $n$. – akkkk Nov 5 '12 at 8:30
@akkkk: $f$ does not depend on $n$, but in order to meet the conditions of the problem, such an $f$ will have particular behavior relative to the sequence of reciprocals of positive integers. Hence one method is to define what $f$ does on each interval $[1/(n+1),1/n]$, making sure the definitions agree at the endpoints. – Jonas Meyer Nov 5 '12 at 8:30