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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 $\mu_1=\mu_2$?

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  • $\begingroup$ Are you acquainted with the uniqueness of measures? $\endgroup$
    – Snoop
    Commented Oct 25, 2022 at 21:28
  • $\begingroup$ @Snoop Yes. My understanding is that this shows that if two Borel measures agree on all open sets, then they agree on all Borel sets. I'm not sure if we can use it to say anything about Lebesgue measurable sets though. $\endgroup$
    – Anonymous
    Commented Oct 25, 2022 at 21:36
  • $\begingroup$ Do you know the result that every Lebesgue measurable set can be written as $B \mathop{\triangle} Z$ where $B$ is Borel and $Z$ has Lebesgue measure zero? If so, then a similar proof shows that each set of Lebesgue measure zero is contained in a Borel set of measure zero. $\endgroup$ Commented Oct 25, 2022 at 21:50

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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, $\mu (E)=\mu (B)=\lambda (B)=\lambda (E)$.

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  • $\begingroup$ Very nice, thanks! Do you know if this proof generalizes? i.e. If $\Sigma$, $\hat{\Sigma}$, $\mu_1$, and $\mu_2$ are as in my question and $E\in\hat{\Sigma}$ do there exist $B,D\in\Sigma$ and $C\in\hat{\Sigma}$ such that $E=B\cup C$, $C\subseteq D$ and $\mu_1(D)=\mu_2(D)=0$? $\endgroup$
    – Anonymous
    Commented Oct 26, 2022 at 1:07

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