MHB Is it enough to show that it is a Lebesgue measure?

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To prove that a translation invariant Borel measure $\mu$ on $\mathbb{R}$ with $\mu([0,1))=1$ is a Lebesgue measure, it is sufficient to demonstrate that $\mu([a,b))=b-a$ for all rational intervals $[a,b)$. The discussion highlights that if two Borel measures coincide on intervals of the form $[a,b)$, they will also agree on the entire $\sigma$-algebra generated by those intervals. This implies that establishing the measure's behavior on rational intervals can extend to all Borel sets. The question raised is whether this is adequate to conclude that $\mu$ is indeed a Lebesgue measure. Ultimately, the focus is on the properties of Borel measures and their implications for Lebesgue measure equivalence.
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Hey! :o

I want to show that when $\mu$ is a Borel measure in $\mathbb{R}$ with $\mu([0,1))=1$, which is a translation invariant, then it is also a Lebesgue measure.

I have shown that $\mu([a,b))=b-a, \forall a,b \in \mathbb{Q}$.

Is it enough to show that $\mu$ is a Lebesgue measure?? (Wondering)
 
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If two Borel measures agree on intervals of the form $[a,b)$ then they agree on anything in the $\sigma$-algebra which is generated by those intervals.
 

Thread 'About the definition of topological manifold using closed sets'

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