Invariant Measure under dilatation We know that the Lebesgue measure is dilatation-invariant, namely $\lambda( \alpha A) = \alpha \lambda(A),$ for any $\alpha > 0$ and a Borel $A$.
What are the conditions for non Lebesgue measure to be dilatation-invariant? Does exist some class of measures having this property?
 A: Partial answer: if $\mu$ is a measure on the sigma algebra of Lebesgue measurable sets in $\mathbb R^{+}$ such that $\mu \{x\}=0$ for al $x$ which has the stated property then there is a constant $c$ such that $\mu =c\lambda$. Proof: if $a >0$ then $\mu (0,a)=a\mu(0,1)=ca$ where $c =\mu (0,1)$. Hence $\mu (a,b) =\mu (a,b)-\mu (0,a)=c(b-a)$ for $b>a>0$. This implies that $\mu =c \lambda$.
If you replace $\mathbb R^{+}$ by $\mathbb R$ the conclusion is $\mu =c_1\lambda_1+c_2\lambda_2$ where $\lambda_1$ and $\lambda_1$ are the restrictions of $\lambda$ to $(0,\infty)$ and $(-\infty,0)$. Conversely any such measure is dilation invariant. 
A: This problem in full generality is a bit challenging. You can show that the atomic measure $\mu$ defined on $\mathbb{R}$ by $\mu(\{x\}) = |x|$ and extended linearly will have this property. Note that this measure is pretty pathological as many otherwise nice sets will have infinite measure.
Kavi has a great answer above (as per usual). I was working on this problem for a paper I am writing and found a different argument that I like a little bit better as it can be adapted more broadly. I will present this for $\mathbb{R}$, but the argument can be extended to $\mathbb{R}^n$. Again, mine is only a partial answer as general measures can be very wild. I will work with measures that are absolutely continuous with respect to the Lebesgue measure. These encapsulate weighed (i.e. Riemann-Stieltjes type) integrals and similar. For practical purposes, these are the most useful.
Let $\lambda$ represent the Lebesgue measure and $\mu \ll \lambda$ be a measure satisfying $\mu(\alpha A) = \alpha \mu(A)$. From absolute continuity with respect to $\lambda$, we have $\mu(A) = \int_A f(x)\,d\lambda(x)$ for some $f\in L^1(\mathbb{R},d\lambda)$ and $f\ge 0$ a.e. From the condition $\mu(\alpha A) = \alpha \mu(A)$, we have
$$ \int_{\alpha A} f(x)\,d\lambda(x) = \alpha \int_A f(x)\,d\lambda(x). $$
Suppose that $A = \{x:|x-c|<r\}$, i.e. an interval centered at $c$ with radius $r$. We will let $r\to 0$ shortly. Dividing through by $\lambda(A)$ and using a clever form of $1$, we have
$$ \frac{\int_{\alpha A} f(x)\,d\lambda(x)}{\lambda(\alpha A)} \frac{\lambda(\alpha A)}{\lambda(A)} = \alpha \frac{\int_A f(x)\,d\lambda(x)}{\lambda(A)} $$
Note that $\lambda(\alpha A) = \alpha \lambda(A)$. Moreover, $\alpha A = \{x:|x-\alpha c| < \alpha r\}$ is centered at $\alpha c$. Letting $r\to 0$ and making use of the Lebesgue differentiation theorem, we have
$$ \alpha f(\alpha c) = \alpha f(c) $$
or in other words, $f(\alpha c) = f(c)$ for all $\alpha > 0$ and almost every $c$. Therefore $f$ is constant a.e. and so $\mu$ is a multiple of the Lebesgue measure.
