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This is actually part (a) question asked previously. However, that person had solved part (a) and so no solution is available. I am trying to prove it myself but am running in to difficulties.

$\textbf{Question}$ (a) Show that there is at most one measure $\nu$ on $\mathcal{B}_{(0,\infty)}$ which satisfies the following conditions:

  1. $\nu((1,e]) = 1$
  2. $\nu(cA) = \nu(A)$ for every $c>0$ and $A \in \mathcal{B}_{(0,\infty)}$.

I think that somehow $(1,e]$ will generate all of the sets in $B_{(0, \infty)}$ but I am having a hard time formalizing the proof.

user7090
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    Do you know about uniqueness of Haar measure? Also, you could try to "transplant" the measure $\mu$ to the whole real line using logarithm/exponential function. This will give you a familiar looking question on the real line. – PhoemueX Dec 10 '14 at 06:35

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To expand on @PhoemueX's comment, one way to go about doing this proof is to use the two following facts:

Fact 1. Let $\mu$ and $\nu$ be measures on $\mathcal B_{(0,\infty)}$ and $T:(0,\infty)\to\mathbb R$ be measurable and bijective (with respect to $\mathcal B_\mathbb R$). Then, $\mu=\nu$ if and only if $\mu\circ T^{-1}=\nu\circ T^{-1}$.

Fact 2. The Lebesgue measure $\lambda$ is the only measure on $\mathcal B_\mathbb R$ such that $\lambda\big((0,1]\big)=1$ and $\lambda(c+A)=\lambda(A)$ for every Borel set $A$ and $c\in\mathbb R$.

Fact 1 is fairly easy to prove. Fact 2 is one of the most fundamental property of the Lebesgue measure. Its demonstration can easily be found in most measure theory textbooks or online (see for instance this math.stackexchange question).

Suppose we define the map $T$ as $T(x)=\ln(x)$ for every $x\in(0,\infty)$ (and thus, $T^{-1}(y)=e^y$ for all $y\in\mathbb R$). Clearly, this is bijective and measurable (since it is continuous).

At this point, it shouldn't be too hard to show that if $\nu$ satisfies conditions 1. and 2. in the statement of your question, then $\mu=\nu\circ T^{-1}$ is such that $\mu\big((0,1]\big)=1$ and that $\mu(c+A)=\mu(A)$ for all $A$ Borel. Then use facts 1 and 2.

Also mentioned by @PhoemueX, this is a special case of the uniqueness of the Haar measure:

Let $G$ be a locally compact group equipped with a topology making the operations of group product and inverse continuous (for example, the set $(0,\infty)$ with the usual multiplication is a group, and we can equip this group with the standard Euclidean topology, making it locally compact, and the multiplication and inverse are continuous). Let $\mathscr B(G)$ be the Borel $\sigma$-algebra on $G$. A measure $\mu$ on $\mathscr B(G)$ is said to be a left Haar measure if $\mu(cA)=\mu(A)$ for every $c\in G$ and $A\in\mathscr B(G)$.

The general proof of the existence and uniqueness of the Haar measure is quite sophisticated, but the proof for $\mathbb R$ equipped with the usual addition (which yields the Lebesgue measure) is much easier and was known for a very long time. In this context, it is interesting to note that the map $T(x)=\ln(x)$ is actually both a group homomorphism and a homeomorphism between $\big((0,\infty),\cdot\big)$ and $(\mathbb R,+)$.

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