Assume $p$ to be a finite measure on $\mathbf R_+$ for the distribution of wealth among a group of people and $\pi$ its normalization to $1$ (ie $\pi(\mathrm d r) = p(\mathrm d r)\left(\int_0^\infty p(\mathrm d s)\right)^{-1}$). Then $T = \int_0^\infty r~p(\mathrm d r)$ is the total wealth of the group, and $M = \int_0^\infty r~\pi(\mathrm d r)$ is the mean wealth of the individuals.
Now for any $q \in [0,1]$, let $Q_\pi(q)$ be the $q$-quantile of the distribution $\pi$, such that $\pi([0,Q_\pi(q)]) = q$. One can define the Lorenz function as for any $q \in [0,1]$ : $$ L(q) = \frac{1}{T} \int_0^{Q_\pi(q)} r~p(\mathrm d r) $$ ie $L(q)$ is the share of wealth detained by que share of $q$ poorest individuals.
Then the Gini index is often defined as the value: $$ G(p) = 1 - 2 \int_0^1 L(q)~\ell(\mathrm d q) $$ $\ell$ being the standard Lebesgue measure on $[0,1]$.
I have also found the Gini index defined as the ratio of the $L^1$-deviation of the wealth by the mean wealth, normalized by a factor of two ie: $$ H(p) = \frac{1}{2M} \int_0^\infty \int_0^\infty |r - s|~\pi(\mathrm d r) \pi(\mathrm d s)$$
But how can one prove that these two definitons are equivalent?
Thanks in advance!
Begin with the remark that $$\int\int|r-s| \pi(dr)\pi(ds) = 2* \int\int_{r\geq s}(r-s) \pi(dr)\pi(ds) =2 \int_{r=0}^\infty \int_{s=0}^r (r-s) \pi(dr)\pi(ds)$$Then $$H(p) =\frac1M\int_{r=0}^\infty r\left( \int_{s=0}^r \pi(ds)\right)\pi(dr)-\frac1M\int_{r=0}^\infty\int_{s=0}^r s\pi(ds)\pi(dr) $$
But from the other side, $$\int_0^1 L(q) dq = \frac1T\int_{q=0}^1 \int_{r=0}^{Q(q)} rp(dr) dq\\ = \int_{r=0}^\infty \frac{r}T\left(\int_{Q^{-1}(r)}^1 dq \right)p(dr)\\\int_{r=0}^\infty \frac{r}T\left(1-Q^{-1}(r) \right)p(dr)$$ And by definition, $Q^{-1}(r) = \int_0^r \pi(ds)$, so
$$\int_0^1L(q)dq=\frac1T\int_0^\infty rp(dr)- \frac1T\int\int_{0\leq s \leq r}rp(dr)\pi(ds)\\ = 1 - \frac1M\int_{0\leq s\leq r} r\pi(dr)\pi(ds)$$
But we can also write $\int_{r=0}^\infty\int_{s=0}^rs\pi(ds)\pi(dr) = \int_{s=0}^\infty s\int_{r=s}^\infty\pi(dr) \pi(ds) = \int_0^\infty s(1-Q^{-1}(s))\pi(ds)= M\int_0^1L(q)dq$ so $$\int_0^1L(q) = \frac1M \int_{r=0}^\infty\int_{s=0}^rs\pi(ds)\pi(dr)$$
By adding the two formulas for $\int_0^1 L(q) dq$, one finds $$2\int_0^1L(q)dq = 1 -\frac1M\int_{0\leq s\leq r} r\pi(dr)\pi(ds)+\frac1M \int_{r=0}^\infty\int_{s=0}^rs\pi(ds)\pi(dr)\\ = 1 -\frac1M\int_{0\leq s\leq r}(s-r)\pi(dr)\pi(ds)$$
Which gives the expected formula.