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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!

Jose Avilez
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MSA2016
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    One may consider $\ln : (0, \infty) \to \mathbb R$ and the push forward $\rho (B) = \mu (\ln^{-1}(B))$. Then $\rho$ is translation invariant and this is well known to be the lebesgue measure (see here) – Arctic Char Aug 15 '21 at 14:48
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    ^ the linked answer doesn't indicate the most technical part of the proof, which is going from intervals to all Borel/Lebesgue measurable sets. This requires a uniqueness of extension argument (see for example Folland's Theorem 1.14). Another common way to argue the uniqueness of extension is to consider the collection of Borel sets where the two measures agree and show it is a $\sigma$-algebra, and show it contains all intervals (thus the two measures agree on the full Borel $\sigma$-algebra). This can be argued using either the monotone class lemma or Dynkin's $\pi$-$\lambda$ theorem – peek-a-boo Aug 15 '21 at 15:31
  • For the argument using the $\pi$-$\lambda$ theorem, see this set of notes (the argument there can be easily reworded to invoke the monotone class lemma instead if that's something you know). – peek-a-boo Aug 15 '21 at 15:32

1 Answers1

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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.

Jose Avilez
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