Proof that this is/is not a measure?

[SlipEternal]

Let \(\displaystyle X = \mathcal{P}\left( (0,1) \right)\) (the power set of the open unit interval) be a poset under the partial order \(\displaystyle \subseteq\).
Let \(\displaystyle A = \left\{\mathcal{C} \subseteq X \mid \mathcal{C}\text{ is a maximal chain in }X\right\}\)
Let \(\displaystyle f:[0,1] \to X\) be defined \(\displaystyle f(t) := (0,t)\).
Let \(\displaystyle S_{(0,1)}\) be the set of all bijections from \(\displaystyle (0,1)\) to itself.
For each \(\displaystyle \sigma \in S_{(0,1)}\), define \(\displaystyle f_\sigma:[0,1] \to X\) as \(\displaystyle f_\sigma(t) := \{\sigma(x) \mid x \in f(t)\}\).
For notation, let \(\displaystyle I = [0,1], f_\sigma(I):=\{f_\sigma(t) \mid t \in [0,1]\}\).
Let \(\displaystyle \mathcal{B}_I = \{B \subseteq I \mid B\text{ is a Borel set}\}\).
For these purposes, \(\displaystyle m\) is standard Lebesgue measure.
For each \(\displaystyle \mathcal{C} \in A\), let \(\displaystyle S_\mathcal{C} = \) \(\displaystyle \left\{\sigma \in S_{(0,1)} \mid f_\sigma(I) = \mathcal{C}\text{ and }\forall B \in (\mathcal{C} \cap \mathcal{B}_I)\text{ s.t. }m(B)>0, f_\sigma(m(B)) = B\right\}\)
Let \(\displaystyle S_{(0,1)}^* = \bigcup_{\mathcal{C} \in A}{S_\mathcal{C}}\).
For each \(\displaystyle K \subset (0,1)\), let \(\displaystyle S_K = \left\{\sigma \in S_{(0,1)}^* \mid K \in f_\sigma(I)\right\}\) and also, let \(\displaystyle M_K = \left\{t\in [0,1] \mid f_\sigma(t) = K, \sigma \in S_K\right\}\).

Finally, define \(\displaystyle \mu:X \to [0,1]\) as \(\displaystyle \mu(K) := \frac{1}{2}(\inf{M_K} + \sup{M_K})\)

Is it possible to determine if \(\displaystyle \mu\) is a measure? Or if it is even well-defined?

 

[SlipEternal]

Never mind. This problem is not important. It turns out that it is tending to a far easier problem that I can figure out on my own.

Let \(\displaystyle \mathcal{B}\) be the set of Borel sets (all of them in the reals).
For every \(\displaystyle X \subseteq \mathbb{R}\), define \(\displaystyle L_X = \left\{B\in \mathcal{B} \mid B \subseteq X\right\}\). Define \(\displaystyle U_X = \left\{B \in \mathcal{B} \mid X \subseteq B\right\}\). Because \(\displaystyle \emptyset,\mathbb{R} \in \mathcal{B}\) and \(\displaystyle \emptyset \subseteq X \subseteq \mathbb{R}\), both \(\displaystyle L_X\) and \(\displaystyle U_X\) are nonempty. The set \(\displaystyle L_X\) can be thought of as the lower bounds for \(\displaystyle X\) while \(\displaystyle U_X\) can be thought of as upper bounds for \(\displaystyle X\) (since every \(\displaystyle L\in L_X\) has the property that for any \(\displaystyle U \in U_X, L \subseteq X \subseteq U\)).

Define \(\displaystyle \mu:\mathbb{R} \to [0,\infty]\) as \(\displaystyle \mu(X) := \frac{1}{2}(\inf\{m(U) \mid U \in U_X\}+\sup\{m(L) \mid L \in L_X\})\).
Again, \(\displaystyle m\) is standard Lebesgue measure.

Clearly, \(\displaystyle \mu\) is not a measure (although I haven't gone through the proof of why).