# When do two measures coincide?

To prove the below statement, I face difficulties to get the meaning of the term 'coincide' in the context of measures.

Let $(S, \Sigma, \mu)$ be a measure space. Denote by $\mathcal{N}$ the collection of all $(\mu, \Sigma)$-null sets. Define \begin{align} \Sigma^* = \{ E \subseteq S,\ \exists F,G \in \Sigma\ \text{such that } F \subseteq E \subseteq G\ \text{ and } \mu(G \setminus F) = 0 \}. \end{align} For $E \in \Sigma^*$ and $F,G$ as above, we define $\mu^*(E) = \mu(F)$. $(•)$

Now, I want to show that $\mu^*$ restricted to $\Sigma$ coincides with $\mu$.

I would formulate the restriction of $\mu^*$to $\Sigma$ as follows: $\mu^*(A),\ A \in \Sigma$.

It could be shown that $\Sigma^*$ is a $\sigma$-algebra and that $\Sigma^* = \sigma( \mathcal{N} \cup \Sigma )$. Since, \begin{align} \Sigma &\subseteq \Sigma \cup \mathcal{N} \\ \Sigma &\subseteq \sigma( \Sigma \cup \mathcal{N} ) = \Sigma^*, \end{align} it follows that $\forall E \in \Sigma \subseteq \Sigma^*: \mu^*(E) = \mu(F)$ as defined in $(•)$. Hence, $\mu$ and $\mu^*$ coincide on $\Sigma$.

Is this the correct interpretation of coinciding measures?

So far what you have proven is that $\Sigma \subseteq \Sigma^*$. This is a good start. Now what you need to prove is that if $E \in \Sigma$, then $\mu^*(E) = \mu(E)$; this is meaningful, because as $\Sigma \subseteq \Sigma^*$, then $A$ is also in $\Sigma^*$. "Coincide" is another word for "being equal".
At some point you have probably proved that $\mu^*(E)$ did not depend on the choice of $F,G \in \Sigma$ such that $\mu(G \setminus F) = 0$. So now take $E \in \Sigma$; to compute $\mu^*(E)$, you know you can choose whichever $F,G \in \Sigma$ such that $F \subseteq E \subseteq G$ and $\mu(G \setminus F) = 0$.
There is an obvious choice: since $E \in \Sigma$, you can take $F = G = E$ itself! It is indeed true that $E \subseteq E \subseteq E$, and that $\mu(E \setminus E) = 0$. Your definition of $\mu_*(E)$ then tells you that $\mu_*(E) = \mu(F)$, but $F = E$, so $\mu_*(E) = \mu(E)$. This is exactly what you wanted to prove.