# probability theory – Prove that the outer measure is greater than or equal to the inner measure

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{ sumlimits_{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{ sumlimits_{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 $$mu_*(A) le mu^*(A)$$?

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