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

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The discussion centers on the properties of Borel measures in relation to Lebesgue measures. Specifically, it establishes that if a Borel measure $\mu$ on $\mathbb{R}$ satisfies $\mu([0,1))=1$ and is translation invariant, then it can be concluded that $\mu$ is indeed a Lebesgue measure. The participant has demonstrated that $\mu([a,b))=b-a$ for all rational intervals $[a,b)$, which is a critical step in proving the equivalence. The consensus is that agreement on intervals of the form $[a,b)$ implies agreement across the entire generated $\sigma$-algebra.

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  • Understanding of Borel measures and their properties
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mathmari
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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.
 

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