Premeasures on Algebras

Question

I have a question that I can't seem to work out. Specifically, I don't entirely understand what is being asked.

The questions is:

Let $\mu_0$ be a premeasure on an algebra $\mathcal{A}$. Let $\mu^*$ be the outer measure induced from $\mu_0$ and let $\mu = \mu^*\mid_{\mathcal{M(A)}}.$ Let $\theta$ be the outer measure induced from $\mu$. We let $\mathcal{M}^*$ denote the $\sigma-$algebra of splitting sets for the outer measure $\mu^*$ which is not (yet) known to coincide with the $\sigma-$algebra of splitting sets of $\theta$.

a.) Prove that $\theta =\mu^*.$

b.) Suppose that $\mu_0$ is $\sigma-$finite. Prove that $\mu^*\mid_{\mathcal{M^*}}$ equals the completion of $\mu$.

I don't entirely understand what the question is asking. Does $\mathcal{M(A)}$ denote the splitting sets of $\mu^*$? So if $\theta(E) = \inf\lbrace \mu^*(A): E\subseteq A; A\in \mathcal{M(A)}\rbrace$, are we looking for the sets that satisfy the splitting condition for $\theta?$ Presumably these would align with $\mathcal{M(A)}$ correct? Would this be enough to prove the first part? I'm not really sure how to do the second part.

Thanks!

Answer 1

The set $\mathcal{M(A)}$ denotes the $\sigma$-algebra generated by $\mathcal A$. There is a result which assures that $\mu^*|_\mathcal{M(A)}$ is a measure.

The point of the exercise is to generalize the equivalence of three different approachs for the construction of the Lebesgue measure, that we list below:

(1) Consider the premeasure $m$, defined on the algebra $\mathcal I$ of the intervals of $\mathbb R$, by $$m([a,b])=m((a,b])=m([a,b))=m((a,b))=b-a,$$ then consider the outer measure $m^*$ generated by $m$ and define the Lebesgue measure to be the restriction $m^*|_{\mathcal M^*}$, where $\mathcal M^*$ denotes the $\sigma$-algebra of splitting sets for $m^*$.

We have that $\mu=m^*|_{\mathcal{M(I)}}$ is a measure (this fact holds for any premeasure). Note that $\mathcal{M(I)}$ is the $\sigma$-algebra of the borelian sets and that $\mu$ is the Jordan measure. The other two approaches are:

(2) Consider the outer measure $\mu^*$, generated by $\mu$, and define the Lebesgue measure to be the restriction $\mu^*|_{\mathcal N^*}$, where $\mathcal N^*$ denotes the $\sigma$-algebra of splitting sets for $\mu^*$.

(3) Define the Lebesgue measure to be the completion of $\mu$.

By the item $a)$ of the exercise, that $\mu^*=m^*$, then so it does $\mathcal N^*=\mathcal M^*$, and consequently, $\mu^*|_{\mathcal N^*}=m^*|_{\mathcal M^*}$, which proves the equivalence of $(1)$ and $(2)$. Then, item $b)$ proves that $(3)$ is equivalent to $(1)$.

The exercise proves that these three approaches are equivalent for any premeasure in any algebra, given that the premeasure is $\sigma$-finite.

Let us prove item a) of the exercise, then item b) follows from item a) and the fact that the restriction of $\theta$ to the $\sigma$-algebra of the splitting sets for $\theta$ is the completion of $\mu$.

Suppose that $\mathcal A$ is an algebra in $X$. For any $A\in \mathcal A$, we have that $A\in\mathcal{M(A)}$ and $\mu_0(A)=\mu^*(A)$. Fix $E\subset X$. For any $(A_n)_{n\in\mathbb N}\subset \mathcal A$ covering $E$, we have, in particular, that $(A_n)_{n\in\mathbb N}\subset \mathcal{M(A)}$, then, by the definition of $\theta$, $$ \theta(E) \leq \sum_{n=1}^\infty \mu^*(A_n) = \sum_{n=1}^\infty \mu_0(A_n). $$ Then $$ \theta(E) \leq \inf\left\{\sum_{n=1}^\infty \mu_0(A_n) : \mbox{$(A_n)_{n\in\mathbb N}\subset \mathcal A$ covers $E$}\right\} = \mu^*(E). $$ It lasts to prove that $\mu^*(E)\leq \theta(E)$.

Fix $\varepsilon>0$ and pick $(A_n)_{n\in\mathbb N}\subset \mathcal{M(A)}$ covering $E$ such that $$ \sum_{n=1}^\infty \mu^*(A_n)\leq \theta(E)+\frac{\varepsilon}{2}. $$ By the definition of $\mu^*$, for each $A_n$, there exists $(B^n_j)_{j\in\mathbb N}\subset \mathcal A$ covering $A_n$ such that $$ \sum_{j=1}^\infty \mu_0(B^n_j)\leq\mu^*(A_n)+\frac{\varepsilon}{2^{n+1}}. $$ Then, $(B^n_j)_{(n,j)\in\mathbb N\times\mathbb N} \subset \mathcal A$ covers $E$, so $$ \mu^*(E)\leq\sum_{(n,j)\in\mathbb N\times\mathbb N} \mu_0(B^n_j) \leq \sum_{n=1}^\infty \mu^*(A_n) + \sum_{n=1}^\infty\frac{\varepsilon}{2^{n+1}} \leq \theta(E) + \varepsilon. $$ Since this is valid for every $\varepsilon>0$, it follows that $\mu^*(E)\leq\theta(E)$.