Borel measure

**Benmath** · August 9th 2010, 03:30 PM

Hi,

I need to prove that their exists a Borel measure v on $\mathbb{R}^+$ which is absolutely continuous with respect to Lebesgue measure and which satisfies $v([a,b])=b^2-a^2$ with $a< b$

I know that the usual Lebesgue measure would satisfy $v([a,b])=b-a$ , but I'm terrible at coming up with examples. If I could find one that satisfies the requirement, I'm sure I could prove it is absolutely continuous. Any hints?

Everything I was trying put me in $\mathbb{R}^2$ , and I need to be in $\mathbb{R}$ . Thanks.

*Note I had a typo and edited that the Lebesque measure satisfies $v([a,b])=b-a$

**Opalg** · August 11th 2010, 07:29 AM

Originally Posted by Benmath

Hi,

I need to prove that their exists a Borel measure v on $\mathbb{R}^+$ which is absolutely continuous with respect to Lebesgue measure and which satisfies $v([a,b])=b^2-a^2$ with $a< b$

I know that the usual Lebesgue measure would satisfy $v([a,b])=b-a$ , but I'm terrible at coming up with examples. If I could find one that satisfies the requirement, I'm sure I could prove it is absolutely continuous. Any hints? [/tex]

It looks as though the measure v should be defined by $v(E) = \int_E2x\,dx$ for each Borel set E, where dx means integral with respect to Lebesgue measure. Is that enough of a hint?

**Benmath** · August 12th 2010, 05:32 PM

Thank you Opalg! That is incredibly obvious! I am usually terrible with the very obvious things. Again, thanks!

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