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If $\displaystyle f(x)=\frac{1}{x^p}$ $(0 < x \leq 1)$ then $f \in L[0,1]$ if $p<1$ and

$$\int_{0}^1 f= \frac{1}{1+p} $$

I know that non negative measurable function f is Lebesgue integrable on [a,b] if

$$\int_{a}^b f=\lim_{n \to \infty} \int_{a}^b f^n$$ If this limit is finite then function is Lebesgue integrable. but how can i find $f^n$ for this function? please help me.Thanks in advance.

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    I'm not sure your question makes a lot of sense. In the case of $L[0,1]$, we normally say that $f\in L[0,1]$ (i.e. is Lebesgue integrable over [$0,1]$) if $\int_0^1 |f|$ is finite.2012-12-01
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    @SamuelReid Here i have to show that $f \in L[0,1]$ for this i'm thinking about some cut of function $f^n$ how can i find such function? and then i have to show the limit of integral of cut-of-function is 1/p+12012-12-01
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    As I said, to show that $f\in L[0,1]$, just calculate $\int_0^1 |f|$ and claim that it is finite. I don't understand what you are trying to do. What is a cut of function?2012-12-01
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    Interesting misprint ... he means $f \wedge n$, the minimum of $f$ and $n$, but when he put that into LaTeX, it came out $f^n$.2013-01-12

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