2
$\begingroup$

The full problem statement is as follows: Let $\mathcal{A}$ be an algebra on X. let let $\mathcal{A}_{\sigma}$ be the set of countable unions in $\mathcal{A}$. Let $u_{0}$ be a premeasure on $\mathcal{A}$ and let $u^*$ be the induced outer measure. Show that for all $E\in X$ and $\epsilon>0$, there exists $A\in \mathcal{A}_{\sigma}$ so that $u^{*}(A) \leq u^*(E) + \epsilon$

SO my initial thought is lets bring u^*(E) to the other side and use triangle inequality to make $u^*(A \setminus E) <\epsilon$. (where I use $A \setminus E$ to mean $A$ without any element of $E$).

now where I'm stuck on is how can I justify that I can pick A arbitrarily close to but not exactly $E$? is it as simple as letting $A = E \cup{\{e\}}$ where $u^*(e) < \epsilon$ ?

$\endgroup$

1 Answer 1

1
$\begingroup$

The definition of $\mu^*$ is as follows:$\mu^*(E)=\inf\{\sum_{j=1}^\infty \mu_0(E_j)|A\subseteq \cup_{j=1}^\infty E_j,E_j\in A\}$.

Therefore for any $E$, $\mu^*(E)+\epsilon > \mu^*(E)$. Hence, $\mu^*(E)+\epsilon$ is larger than the infimum. Therefore by the definition of infimum there exists $\{E_j\}_{j\in N}, E_j\in A$ such that $A\subseteq \cup_{j=1}^\infty E_j$ and $\mu^*(E)+\epsilon >\sum_{j=1}^\infty \mu_0(E_j) $.

Note: $\cup_{j=1}^\infty E_j\in A_\sigma$ and $\mu^*(\cup_{j=1}^\infty E_j)\leq \sum_{j=1}^\infty \mu_0(E_j)$.

Hence, $\mu^*(\cup_{j=1}^\infty E_j)<\mu^*(E)+\epsilon$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.