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Let $a<b$ and $F$ the σ-algebra generated by all intervals in $[a,b]$. Let $μ$ ,$ν$ finite measures in space $([a,b],F)$ such that for all $c∈[a, b]$ we have $μ([a,c])=ν([a,c])$ show that $μ=ν$ .

Did
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Alex Med
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    Has your class covered Dynkin's Lemma? If yes, use that. If no, you essentially have to prove something like it. So can answer assume Dynkin's Lemma? – gnometorule Jan 29 '13 at 06:14
  • Monotone Class Theorem is perhaps a more common name for Dynkins's lemma mentioned by @gnometorule – Julian Wergieluk Jan 29 '13 at 07:39

2 Answers2

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Let $\mathcal C$ denote the class of the measurable subsets $A$ on $[a,b]$ such that $\mu(A)=\nu(A)$ and $\mathcal I$ the class of the intervals $[c,d]$ with $c\leqslant d$ in $[a,b]$. Then:

  1. $\mathcal C$ is a sigma-algebra.
  2. $\mathcal I\subseteq\mathcal C$ by hypothesis.
  3. $\sigma(\mathcal I)=\mathcal B([a,b])$ by definition of $\mathcal B([a,b])$.

Thus, $\mathcal B([a,b])=\sigma(\mathcal I)\subseteq\sigma(\mathcal C)=\mathcal C$. Since $\mathcal C\subseteq\mathcal B([a,b])$ by definition, $\mathcal C=\mathcal B([a,b])$. QED.

Did
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  • The definition of the $\sigma$-algebra in the question is that it is generated by all intervals in $[a,b]$, not just the closed intervals that begin at $a$, which constitute $\mathcal I$. So a little more work is needed to show that all the subintervals of $[a,b]$ can be generated from those in $\mathcal I$ by the operations of $\sigma$-algebras. – Andreas Blass Apr 29 '13 at 14:29
  • @AndreasBlass You mean, a little less work is needed since one starts from a larger collection of intervals... Edited. – Did Apr 29 '13 at 16:30
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  • The property gives that $\mu(I)=\nu(I)$ for all interval $i$ contained in $[a,b]$.
  • Hence equality takes place for finite unions of intervals, which form an algebra $\mathcal A$.
  • Given a measurable set $S\subset [a,b]$, take $S'\in\mathcal A$ such that $\mu(S\Delta S')+\nu(S\Delta S')<\varepsilon$, where $\varepsilon$ is arbitrary but fixed. This is possible since $\mu+\nu$ is a non-negative finite measure. Then $|\mu(S)-\nu(S)|\leqslant\varepsilon$.
Community
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Davide Giraudo
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