Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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. –  Samuel Reid Dec 1 '12 at 4:41
    
@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+1 –  Siddhant Trivedi Dec 1 '12 at 5:12
    
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? –  Samuel Reid Dec 1 '12 at 5:21
1  
Interesting misprint ... he means $f \wedge n$, the minimum of $f$ and $n$, but when he put that into LaTeX, it came out $f^n$. –  GEdgar Jan 12 '13 at 16:39

1 Answer 1

Let $p<1$ and

$$f_n(x) := \frac{1}{x^p} \cdot 1_{\left[\frac{1}{n},1\right]}(x) \qquad (n \in \mathbb{N})$$

(where $1_B$ denotes the indicator function of a set $B$). Then clearly $f_n(x) \uparrow f(x)$ for all $x \in (0,1]$. By monotone convergence we obtain

$$\int_0^1 f(x) \, dx = \sup_{n \in \mathbb{N}} \int_{0}^1 f_n(x) \,dx \left(= \lim_{n \to \infty} \int_0^1 f_n(x) \, dx \right) < \infty \qquad (\ast)$$

(since $f_n$ is bounded on $[0,1]$ (hence $f_n \in L^1$) you can calculate the integral $\int_0^1 f_n$ as usual). Thus $f \in L^1$. From $(\ast)$ you will also obtain the equality you are looking for.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.