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Let $\mu$ be a pre-probability measure on algebra $\mathcal{A}$, and define inner measure $\mu_*: \mathcal{P}(\Omega) \to [0,1]$ and outer measure $\mu^*: \mathcal{P}(\Omega) \to [0,1]$ as:

\begin{align*} \mu_*(A) &= \text{sup} \left\{ \sum\limits_{i=1}^\infty \mu(B_i) : \text{for all disjoint sequences such that} \; B_i \in \mathcal{A}, \; \cup_{i \ge 1}^\infty B_i \subset A \right\} \\ \mu^*(A) &= \text{inf} \left\{ \sum\limits_{i=1}^\infty \mu(C_i) : \text{for all sequences such that} \; C_i \in \mathcal{A}, A \subset \cup_{i=1}^\infty C_i \right\} \\ \end{align*}

Prove that $\mu_*(A) \le \mu^*(A)$ for all $A$

Caveat: $\mathcal{A}$ is an algebra over $\Omega$, so it is closed over finite union, but not countable unions, so countable unions may not exist in $\mathcal{A}$.

We can define $C'_1 = C_1$, $C_{i > 1} = C_i \setminus \cup_{j=1}^{i-1} C_j$ where $C_i'$ are disjoint, $\cup_i C'_i = \cup_i C_i$ and $\sum_i \mu(C'_i) \le \sum_i \mu(C_i)$, so without loss of generality we can assume that $C_i$ are disjoint.

For any such sequences $B_i, C_i$, we have:

\begin{align*} \cup_{i \ge 1}^\infty B_i \subset A \subset \cup_{i=1}^\infty C_i \\ \end{align*}

With monotonicity of both the inner and outer measure:

\begin{align*} \mu_*(\cup_{i \ge 1}^\infty B_i) &\le \mu_*(A) \le \mu_*(\cup_{i=1}^\infty C_i) \\ \mu^*(\cup_{i \ge 1}^\infty B_i) &\le \mu^*(A) \le \mu^*(\cup_{i=1}^\infty C_i) \\ \end{align*}

Here, I'm somewhat stuck. How do I show that $\mu_*(A) \le \mu^*(A)$?

clay
  • 2,883

1 Answers1

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Recall that $\mu (A)=\mu^{*}(A)$ if $A \in \mathcal A$ and that $\mu^{*}$ is monotone.

Whenever $B_i,C_i \in \mathcal A$, $B_i's$ are disjoint, $\bigcup B_i \subset A$ and $A \subset \bigcup C_i$ we have $\sum\limits_{k=1}^{n} \mu(B_i) =\mu (\bigcup_{k=1}^{n} B_i) =\mu^{*} (\bigcup_{k=1}^{n} B_i)\leq \mu^{*} (\bigcup_{k=1}^{\infty} C_i)\leq \sum\limits_{k=1}^{\infty} \mu(C_i) $ for each $n$. Letting $n \to \infty$ we get $\sum\limits_{k=1}^{\infty} \mu(B_i) \leq \sum\limits_{k=1}^{\infty} \mu(C_i) $. The proof is now finished by taking infimum over all $(C_i)$ and supremum over all $(B_i)$.

Kavi Rama Murthy
  • 359,332