Function that Satisfies Axioms of Uncertainty/Proof

Proof
Let $g: \Z \to \R$ be the mapping defined as:


 * $(1): \quad \map g n = \map H {\dfrac 1 n, \dfrac 1 n, \dotsc, \dfrac 1 n}$

Let $k \in \Z_{>0}$.

We have:

Let $r, s \in \R$ and $n \in \Z_{>0}$.

Let $m$ satisfy:
 * $(3): \quad r^m \le s^n \le r^{m + 1}$

From $(2)$ and Axiom 5:


 * $\map g {r^m} \le \map g {s^n} \le \map g {r^{m + 1} }$

and so:


 * $m \map g r \le n \map g s \le \paren {m + 1} \map g r$

Taking logarithms of $(3)$:
 * $m \ln r \le n \ln s \le \paren {m + 1} \ln r$

which holds because Logarithm is Strictly Increasing.

Hence:
 * $\size {\dfrac {\map g s} {\map g r} - \dfrac {\map \ln s} {\map \ln r} } \le \dfrac 1 n$

We have that $n$ is an arbitrary positive integer.

Hence the of the above can be made arbitrarily small.

Hence:
 * $\size {\dfrac {\map g s} {\map g r} - \dfrac {\map \ln s} {\map \ln r} } = 0$

and so:
 * $(4): \quad \dfrac {\map g s} {\map \ln s} = \dfrac {\map g r} {\map \ln r} = A$

for some constant $A$.

That is:
 * $(5): \quad \map g s = A \map \ln s$

for $s \in \Z$.

Let $p$ be a rational number such that $0 < p < 1$.

That is:
 * $p = \dfrac t n$

for some integers $t$ and $n$.

Thus we can set:
 * $q = 1 = p = \dfrac {n - t} n$

Axiom 8 gives:


 * $\map g n = \map H {\dfrac 1 n, \dfrac 1 n, \dotsc, \dfrac 1 n} = \map H {\dfrac 1 n, \dfrac {n - t} n} + \dfrac t n \map g t + \dfrac {n - t} n \map g {n - t}$

Using $(5)$ and gathering up terms:


 * $\map H {\dfrac 1 n, \dfrac {n - t} n} = A \paren {\dfrac t n} \ln \dfrac t n + A \paren {\dfrac {n - t} n} \ln \dfrac {n - t} n$

which gives:


 * $(6): \quad \map H {p, 1 - p} = A p \ln p + A \paren {1 - p} \, \map \ln {1 - p}$

for rational $p$.

Axiom 6 gives that $H$ is continuous.

Thus $(6)$ extends to all real $p$ such that $0 < p < 1$.

It remains to be demonstrated that:
 * $(7): \quad \map H {p_1, p_2, \ldots, p_N} = A \displaystyle \sum_{i \mathop = 1}^N p_i \ln p_i$

where:
 * $0 < p_i < 1$ for all $p_i$
 * $p_1 + p_2 + \dotsb + p_N$

for all $N \in \Z_{\ge 2}$.

We have demonstrated that $(7)$ is true for $N = 2$.

Assume the induction hypothesis that $(7)$ holds for $N = k$.

Consider:
 * $\map H {p_1, p_2, \ldots, p_k, p_{k + 1} }$

Let:
 * $p = p_1 + p_2 + \dotsb + p_k$
 * $q = p_{k + 1}$

and apply Axiom 8:

Hence the result from the Principle of Mathematical Induction.