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 $μ=ν$ .
 A: 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:


*

*$\mathcal C$ is a sigma-algebra.

*$\mathcal I\subseteq\mathcal C$ by hypothesis. 

*$\sigma(\mathcal I)=\mathcal B([a,b])$ 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.
A: *

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

