There is a unique Borel measure on $(0,\infty)$ with scaling invariance and $\mu([1,e])=1$. 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!
 A: 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.
