Suppose that $\mu$ is a Borel measure on $(0,\infty)$ with $\mu([1,e])=1$ and for all $c>0$ and all $A$ Borel measurable, we have $\mu(cA)=\mu(A)$. Show that there exists a unique $\mu$ with these properties.
It is clear to me that this measure is just $\mu(A)=\int_A \frac{1}{x}\,\text{d}x$. Both of the properties are easy to check. However I am confused about uniqueness.
I have tried supposing that there are $\mu_1,\mu_2$ with these properties and considering the signed measure $\mu_1-\mu_2$. I also have thought about using a Lebesgue decomposition, arguing that the singular part must be identically the zero measure, and then reducing to the case when $\mu$ is also absolutely continuous with respect to Lebesgue measure. Overall I feel like I can't really do much with the scaling invariance. The algebra is just not being nice to me today.
Any hints will help!
This argument is essentially the one given in Theorem 11.9 in Folland's Real Analysis; there he proves translation invariance for general Haar measures on Abelian groups.
Theorem If $\mu$ and $\nu$ are measures on $(0,\infty)$ with the scaling invariance above, then $\mu = c \nu$ for some $c > 0$.
Proof Let $h \geq 0$ be a positive continuous compactly supported function on $(0, \infty)$ which is inversion invariant: $h(x) = h(x^{-1})$. You can construct such a function by setting $h(x) = g(x) + g(x^{-1})$.
Then, for any continuous compactly supported $f$, you get:
$$\begin{align*} \int h d \nu \int f d \mu &= \int \int h(y) f(x) d \mu (x) d \nu (y) = \int \int h(y) f(xy) d \mu (x) d \nu (y) \\ &= \int \int h(x^{-1}y) f(y) d \nu (y) d \mu(x) = \int \int h(xy^{-1}) f(y) d \nu (y) d \mu(x) \\ &= \int \int h(xy^{-1}) f(y) d \mu(x) d \nu (y) = \int \int h(x) f(y) d \mu(x) d \nu (y) \\ &= \int h d \mu \int f d \nu \end{align*}$$
Where we have used Tonelli's theorem and change-of-variables extensively. Setting $c = \frac{\int h d \mu}{\int h d \nu}$ and using inner regularity we conclude that $\mu = c \nu$. $\blacksquare$
In your situation, your measure is already normalised, so you end up with $c=1$, which gives uniqueness.