Lemma 4

1° Let $f$ be a lower semi-continuous and bounded positive function on $T$. Then its class $\dot{f}$ in $L^\infty$ is the supremum of the set of classes $\dot{\varphi}$, where $\varphi$ runs over the set of functions in $\mathscr{K}(T)$ such that $0 \leq \varphi \leq f$.

2° Let $f$ be a measurable and bounded positive function on $T$. Then its class $\dot{f}$ in $L^\infty$ is the infimum of the set of classes $\dot{\psi}$, where $\psi$ runs over the set of lower semi-continuous and bounded functions on $T$ that are $\geq f$.

1° Let $f'$ be a function in $\mathscr{L}^\infty$ such that $\dot{f}'$ is the supremum in $L^\infty$ of the set of classes $\dot{\varphi}$ of the functions $\varphi$ in $\mathscr{K}(T)$ such that $0 \leq \varphi \leq f$; obviously $\dot{f}' \leq \dot{f}$. Let $U$ be a relatively compact open subset of $T$; for every function $h$ in $\mathscr{K}(T)$ such that $0 \leq h \leq f \varphi_U$ one has, by definition, $h(t) \leq f'(t)$ locally almost everywhere, therefore $h(t) \leq f'(t) \varphi_U(t)$ almost everywhere; it follows that $\int h \, d\mu \leq \int f' \varphi_U \, d\mu$. However, since $f \varphi_U$ is lower semi-continuous, $\int f \varphi_U \, d\mu = \sup \int h \, d\mu$, where $h$ runs over the set of functions in $\mathscr{K}(T)$ such that $0 \leq h \leq f \varphi_U$ (Ch. IV, §1, No. 1, Def. 1); therefore

and since $f' \varphi_U \leq f \varphi_U$ almost everywhere, necessarily $f \varphi_U = f' \varphi_U$ almost everywhere, whence $f = f'$ locally almost everywhere.

2° Let $f'$ be a function in $\mathscr{L}^\infty$ such that $\dot{f}'$ is the infimum in $L^\infty$ of the set of classes $\dot{\psi}$ of the lower semi-continuous functions $\psi$ that are bounded and $\geq f$; then $\dot{f}' \geq \dot{f}$. Let $K$ be a compact subset of $T$; for every lower semi-continuous function $h$ that is bounded and $\geq f \varphi_K$, let $\overline{h}$ be the function equal to $h$ on $K$ and to $\|f\| + \|h\|$ on $T - K$. Then $\overline{h}$ is lower semi-continuous and $\geq f$, therefore by definition $\overline{h}(t) \geq f'(t)$ locally almost everywhere; it follows that $h(t) \geq f'(t)\varphi_K(t)$ almost everywhere, whence $\int h d\mu \geq \int f'\varphi_K d\mu$. But $\int f\varphi_K d\mu = \inf \int h d\mu$, where $h$ runs over the set of lower semi-continuous functions that are bounded and $\geq f\varphi_K$ (Ch. IV, §1, No. 3, Def. 3); therefore

and since $f\varphi_K \leq f'\varphi_K$ almost everywhere, necessarily $f\varphi_K = f'\varphi_K$ almost everywhere, whence $f = f'$ locally almost everywhere.

The lemma having been proved, let $\theta$ be a positive linear form on $\mathscr{L}^\infty$ satisfying the condition in the statement of Prop. 14. The restriction of $\theta$ to the space $\mathscr{K}(T)$ is a positive measure $\nu$ on $T$. We are going to show that, for every positive function $f \in \mathscr{L}^\infty(T, \mu)$, one has $\theta(f) = \nu^*(f)$. Suppose first that $f$ is lower semi-continuous (and bounded); by Lemma 4, $f$ is the supremum of the increasing directed set of classes $\dot{\varphi}$, where $\varphi$ runs over the directed set $\Phi$ of functions in $\mathscr{K}(T)$ such that $0 \leq \varphi \leq f$. Since by hypothesis $\theta(f) = \sup_{\varphi \in \Phi} \theta(\varphi)$, and $\nu^*(f) = \sup_{\varphi \in \Phi} \nu(\varphi)$ by definition, our assertion is proved in this case. Secondly, suppose that $f$ is $\mu$-measurable and bounded; then, by definition, $\nu^*(f) = \inf_{\psi \in \Psi} \nu^*(\psi)$, where $\psi$ runs over the decreasing directed set $\Psi$ of lower semi-continuous functions that are bounded and $\geq f$. If $a \geq \|f\|$ then, applying the hypothesis of the statement to the increasing directed set of classes of the functions $a - \psi$, where $\psi \in \Psi$ and $\psi \leq a$, one sees, by virtue of the lemma, that $\theta(f) = \inf_{\psi \in \Psi} \theta(\psi)$, therefore indeed $\theta(f) = \nu^*(f)$. In particular, for every $\mu$-negligible function $f \geq 0$, one has $\theta(f) = 0$, therefore $\nu^*(f) = 0$ and consequently (No. 5, Th. 2) $\nu$ is a measure with base $\mu$; moreover, $\nu^*(1) = \theta(1) < +\infty$, consequently (Cor. of Th. 1) $\nu = g \cdot \mu$ with $g \in \mathscr{L}^1(T, \mu)$. Finally, since every $\mu$-measurable function is $\nu$-measurable, every positive function $f \in \mathscr{L}^\infty(T, \mu)$ is $\nu$-integrable and $\int fg d\mu = \nu^*(f) = \theta(f)$, which completes the proof.

One concludes from Prop. 14 that the linear forms on $\mathscr{L}^\infty$ of the type $f \mapsto \int fg d\mu$, where $g \in \mathscr{L}^1$, are the differences $\theta_1 - \theta_2$, where $\theta_1$ and $\theta_2$ are positive linear forms satisfying the condition of Prop. 14.