# Are there any measurable sets which are not inside the sigma algebra generated by the algebra $\mathcal{A}$?

I am reading Folland's Real Analysis. I have a doubt about how the following two theorems are related with each other.

Theorem 1.11 (Caratheodory Theorem):

Let $\mu^{*}$ be an outer measure on $X$. Then the set of all $\mu^{*}$ measurable sets form a $\sigma$-algebra $\mathcal{M}$. The restriction of $\mu^{*}$ to $\mathcal{M}$ is a complete measure.

Theorem 1.14:

Let $\mathcal{A}\subseteq P(X)$ be an algebra and $\mu_0:\mathcal{A}\rightarrow [0,\infty]$ be a premeasure. Let $\mathcal{M}$ be the $\sigma$-algebra generated by $\mathcal{A}$. Let $\mu^{*}$ be the outer measure generated from $\mu_0$. Then $\mu^{*}$ defines a measure on $\mathcal{M}$ and the restriction of $\mu^{*}$ to $\mathcal{A}$ is $\mu_0$.

My question is this:

Let us start with a space $X$. Let $\mathcal{A}\subseteq P(X)$ be an algebra and $\mu_0:\mathcal{A}\rightarrow [0,\infty]$ be a premeasure. Let $\mu^{*}$ be the outer measure generated from $\mu_0$. Let $\mathcal{M}_1$ be the $\sigma$-algebra generated by $\mathcal{A}$ and $\mathcal{M}_2$ be the $\sigma$-algebra of all $\mu^{*}$ measurable sets.

(1) It is clear that $\mathcal{M}_1\subseteq \mathcal{M}_2$. But is it true that $\mathcal{M}_1=\mathcal{M}_2$?

(2) Is the restriction of $\mu^{*}$ to $\mathcal{M}_1$ a complete measure?

$\mathcal{A}$ can be the set of finite union of intervals of $\mathbb{R}$ with the usual premeasure( something needs to be proved - countable additivity ). The $\sigma$ algebra generated by $\mathcal{A}$ is the family of Borel subsets. But $\mathcal{M}_2$, the family of Lebesgue-measurable subsets, is larger. It is in fact the completion of $\mathcal{M}_1$ with respect to $\mu$.
I guess a more interesting question is : can $\mathcal{M}_2$ be larger than the completion of $\mathcal{M}_1$? I don't have an answer to that at the moment.