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Does anyone can help me to calculate a Lebesgue integral with the help of the following definition?

$\int f d \mu = \int_0^{\infty} \mu(f^{-1}((t,\infty]))dt \in [0,\infty]$.

The integral to calculate is the following: $\int_{B_1(0)} \vert x \vert^s d \mu(x)$ where $\mu$ denotes the n-dimensional Lebesgue measure and $B_1(0)$ the unit ball and $s \in \mathbb{R}$.

Is it possible here to work with the definition above?

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  • $\begingroup$ First identify the set $f^{-1}((t,+\infty])$, then determine its measure, then integrate. $\endgroup$ – Daniel Fischer Nov 9 '16 at 15:51
  • $\begingroup$ sorry but I am a bit stuck in indentifying the set $f^{-1}((t,+\infty])$. do I need a case by base observation? $\endgroup$ – tubmaster Nov 9 '16 at 16:16
  • $\begingroup$ Yes. First you need to distinguish the cases $s < 0,\, s = 0,\, s > 0$, and then $t = 1$ is a point where the behaviour changes. $\endgroup$ – Daniel Fischer Nov 9 '16 at 16:24
  • $\begingroup$ do you mean the following? for example for s=0: $f^{-1}((t,+\infty]) = \{x: x<-t $ or $ x>t\}$. I am really stuck... :( $\endgroup$ – tubmaster Nov 9 '16 at 16:30
  • $\begingroup$ $f^{-1}((t,+\infty]) = \{ x \in B_1(0) : \lvert x\rvert^s > t\}$. Can you describe that set in words? $\endgroup$ – Daniel Fischer Nov 9 '16 at 16:36
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Ok I hope that I am now on a right way:

In order to usw the definition of the Lebesgue integral above I determined $f^{-1}((t,+\infty])$ as $\{ x \in B_1(0) : \lvert x\rvert^s > t\}$. But in a next step, how can I get the measure of this set, i.e. $\mu(\{ x \in B_1(0) : \lvert x\rvert^s > t\})$?

Any suggestions :)?

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