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