# see when two measures are equal on σ-algebra generated by all intervals in [a,b].

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 $μ=ν$ .

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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 6:14
Monotone Class Theorem is perhaps a more common name for Dynkins's lemma mentioned by @gnometorule –  Julian Wergieluk Jan 29 '13 at 7:39

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.

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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
• 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 $|\mu(S)-\nu(S)|\leqslant\varepsilon$.