Extending Positive Functionals: Linearity How does regularity provide linearity?

Given the full Banach space of bounded functions over a suitable set:
$$\mathcal{B}:=\{f:\Omega\to \mathbb{C}:\|f\|_\Omega&lt;\infty\}$$
and a linear subspace $\mathcal{C}\leq\mathcal{B}$.

Consider a linear functional on the subspace:
$$I:\mathcal{C}\to\mathbb{C}:f\mapsto I(f)$$
that is bounded and positive:
$$|I(f)|\leq \|I\|\cdot\|f\|_\Omega$$
$$I(f)\geq 0\quad(f\geq 0)$$

Extend it formally to all bounded functions by splitting these:
$$g=g_{\Re_+}+g_{\Re_-}+ig_{\Im_+}+ig_{\Im_-}$$
and then defining:
$$I_E(g):=\sup_{f\in\mathcal{C}:f\leq g} I(f)\quad(g\geq 0)$$
(Note that $\sup_{f\leq g}I(f)=\sup_{0\leq f\leq g}I(f)$.)

This gives a sharp estimate for the bound:
$$|I_E(g)|=|\sup_{0\leq f\leq g}I(f)|=\sup_{0\leq f\leq g}I(f)\leq\sup_{0\leq f\leq g}\|I\|\cdot\|f\|_\Omega=\|I\|\cdot\|g\|_\Omega\quad(g\geq 0)$$
and therefore:
$$|I_E(g)|=\Re_+ e^{-i\phi}I_E(g)=I_E(\Re_+ e^{-i\phi}g)\leq \|I\|\cdot\|\Re_+ e^{-i\phi}g\|_\Omega\leq\|I\|\cdot\|g\|_\Omega\quad(re^{i\phi}:=I_E(g))$$
(Note, linearity has not been used yet in full!)

Moreover it remains positive:
$$I_E(g)\geq 0\quad(g\geq 0)$$

Now, it requires additional regularity assumptions in order to maintain linearity...

But which and how to proceed then?



As an example to see what actually can happen consider the integral of simple functions:
$$I:\mathcal{S}([0,1])\to\mathbb{C}:f\mapsto\int_{[0,1]} f\mathrm{d}x$$

and the characteristic functions:
$$\chi_{A},:\mathbb{[0,1]}\to\mathbb{C}$$
$$\chi_{A^c}:\mathbb{[0,1]}\to\mathbb{C}$$
for some nonmeasurable set $A\subseteq[0,1]$.

Then the extension gives:
$$I_E(\chi_A+\chi_{A^c})=\mu([0,1])$$
$$I_E(\chi_A)+I(\chi_{A^c})=\mu^*(A)+\mu^*(A^c)$$
but $A$ was nonmeasurable so the integral fails to be linear here.
You have defined $I$ on the nonnegative functions $g$, then extended to arbitrary $f$ by $I(g) = I(g_{\Re_+})- I(-g_{\Re_-})+i I(g_{\Im_+}) - i I(-g_{\Im_-})$ where $g=g_{\Re_+}+g_{\Re_-}+ig_{\Im_+}+ig_{\Im_-}$ and $g_{\Re_+} \ge 0, g_{\Re_-} \le 0, g_{\Im_+} \ge 0, g_{\Im_-} \le 0$.  Now suppose $f = g + h$.
Why should $I(f) = I(g) + I(h)$?

Well, if all these functions were in $\mathcal C$ this would be true because $I$ is linear on $\mathcal C$.  So we hope for an approximation: if for any $\epsilon &gt; 0$ there are
$\tilde{f}, \tilde{g},\tilde{h} \in \mathcal C$ such that $\tilde{f} = \tilde{g} + \tilde{h}$ and $|I(f) - I(\tilde{f})|$, $|I(g) - I(\tilde{g})|$, $|I(h) - I(\tilde{h})| &lt; \epsilon$
then $|I(f) - I(g) - I(h)| &lt; 3 \epsilon$, and since $\epsilon$ is arbitrary that means $I(f) - I(g) - I(h) = 0$.  Regularity (which you didn't actually state) should allow you to find such $\tilde{f}, \tilde{g}, \tilde{h}$.
The problem is that it misses some sort of monotone convergence theorem:
$$0\leq g_n(x)\uparrow g(x)\implies I_E(g_n)\uparrow I_E(g)$$
Then that would suggest to consider all those functions which are approximable:
$$g\in\mathcal{C}_E:\iff\exists f_n\in\mathcal{C}:f_n(x)\to g(x)$$