Let $\lambda$ be the Lebesgue measure on $\mathbb{R}^n$ and suppose $\mu$ is a measure defined on the Lebesgue $\sigma$-algebra such that $\mu(B)=\lambda(B)$ for all Borel sets $B$. Does it follow that $\mu(E)=\lambda(E)$ for any measurable set $E$? I couldn't find any proof of this, so I am starting to believe there may be a counterexample. Perhaps, there are different ways to extend $\lambda$ from the Borel $\sigma$-algebra to all measurable sets using Zorn's lemma.
If this implication turns out to be true, then there is a more general question one could ask. Let $X$ be a set, $\Sigma$ and $\sigma$-algebra on $X$, and $\hat{\Sigma}$ the completion of $\Sigma$. If $\mu_1$ and $\mu_2$ are two measures defined on $\hat{\Sigma}$ such that $\mu_1(B)=\mu_2(B)$ for all $B\in\Sigma$, does this imply $\mu_1=\mu_2$?
Best Answer
If $E$ is Lebesgue measurable then $E=B \cup C$ where $B$ is a Borel set, $C \subseteq D$ and $D$ is a Borel set of Lebesgue measure $0$. Since $\lambda (D)=\mu (D)=0$ we get $\mu (B) \leq \mu (E) \leq \mu (B)+\mu (D)=\mu (B)$. Hence, $\mu (E)=\mu (B)=\lambda (B)=\lambda (E)$.