# What is the definition of $s$-dimensional Lebesgue measure on $R^d$ for $s\ne d$?

On the Appendix C about Hausdorff dimensions and measures of the book Multiparamerer Processes by Khoshnevisan it is written the following

Lemma 1.1.2 For all Borel sets $E\subset R^d$, and for all real numbers $s\geq 0$, $$H_s(E)\geq Leb_s(E).$$

Proof Let us fix an arbitrary $\varepsilon>0$ and find arbitrarily closed $l^\infty$ balls $B_1, B_2...$ whose radii $r_1, r_2...$ are all less or equal to $\varepsilon$. Whenever $E\subset\cup_i B_i$ $$Leb_s(E)\leq\sum_i Leb_s(B_i)=\sum_i(2 r_i)^s$$ (...)

Where $H_s$ is of course the s-dimensional Hausdorff measure on $R^d$.

My question is the following: Do you know which is the definition of $Leb_s$? He refers to it as the $s$-dimensional Lebesgue measure on $R^d$.

I just know the standard definition of Lebesgue measure https://proofwiki.org/wiki/Definition:Lebesgue_Measure

I also checked on the rest of the book to see if there was a more concrete definition of $Leb_s$ and I didn´t find it... probably I missed it!

• I've never seen any such definition, except of course for $s=d$. Possibly if you look at the proof you can figure out what measure he has in mind... – David C. Ullrich Apr 10 '18 at 13:32
• From the proof you can deduce that if you take an $l_{\infty}$ ball $B$ of radii $r$ then $Leb_s(B)=(2r)^s$. – Adrián Hinojosa Calleja Apr 11 '18 at 11:05
• Really? It's not hard to show that there is no such measure if $s\ne d$. – David C. Ullrich Apr 11 '18 at 12:07
• Yes it's quite strange this doesn´t make any sense... I doble checked in the book and there is no reference in about what is $Leb_s$, I thought it could be a typo and it should say $Leb^s$ but even in this case it doesn´t make any sense... I wrote part of the proof in the post. – Adrián Hinojosa Calleja Apr 12 '18 at 20:15

Lemma 1.1.2 For all Borel sets $E\subset \mathbb{R}^d$, and for all real numbers $s\geq d$, $$H_s(E)\geq Leb_d(E).$$