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.

  • $\begingroup$ 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. $\endgroup$ Commented Dec 1, 2012 at 4:41
  • $\begingroup$ @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 $\endgroup$ Commented Dec 1, 2012 at 5:12
  • $\begingroup$ 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? $\endgroup$ Commented Dec 1, 2012 at 5:21
  • 1
    $\begingroup$ Interesting misprint ... he means $f \wedge n$, the minimum of $f$ and $n$, but when he put that into LaTeX, it came out $f^n$. $\endgroup$
    – GEdgar
    Commented Jan 12, 2013 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.


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