Conditional expectation - a functional analytic proof of existence?

Question

$ \newcommand{\E}{\mathbf{E}} \newcommand{\pp}{\mathbb{P}} \newcommand{\R}{\mathbb{R}} \newcommand{\scrF}{\mathscr{F}} \newcommand{\scrG}{\mathscr{G}} $Let $(\Omega,\scrF,\pp)$ be a probability space and let $\scr{G}\subset\scrF$ be a $\sigma$-subalgebra. Let $X$ be a real, integrable $\scrF$-measurable random variable. A conditional expectation of $X$ relative to $\scrG$ is any real, integrable $\scrG$-measurable random variable $Y$ such that $$\forall A\in\scrG,\quad\E[1_AX]=\E[1_AY]$$ Existence is usually proven by invoking the Radon-Nikodym theorem: the correspondence $A\mapsto\E[1_AX]$ defines a signed measure on $\scrG$; it is clearly absolutely continuous with respect to (the restiction to $\scrG$ of) $\pp$, hence is of the form $Y\cdot\pp$ for a uniquely defined $Y\in\ L^1(\scrG)$.

Question: Is there a "functional analytic" proof of this fact?

By this I mean a proof along the lines of

1. give an abstract characterization of the continuous linear maps $L^\infty(\scrG)\to\R$ that are of the form $Z\mapsto\E[YZ]$ for some $Y\in L^1(\scrG)$,
2. show that the continuous linear form $L^\infty(\scrG)\to\R, Z\mapsto\E[ZX]$ satisfies this criterion, and is thus of the form $Z\mapsto\E[ZY]$ for a uniquely determined $Y\in L^1(\scrG)$.

Another way of formulating the first question is "how does one recognize continuous linear forms on $L^\infty(\scrG)$ that arise from the canonical embedding $L^1(\scrG)\to L^1(\scrG)''\simeq L^\infty(\scrG)'$ under the identification $L^\infty(\scrG)\simeq L^1(\scrG)'$, where $V'$ is the topological dual of the topological vector space $V$.

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I'm wondering whether the Radon-Nikodym theorem is a secretly a result characterizing those continuous linear forms on $L^\infty(\scrG)$ that arise from the embedding $L^1(\scrG)\hookrightarrow L^\infty(\scrG)'$. The characterization of absolutely continuity of measures $\mu$ through $\pp(A)=0\implies\mu(A)=0$ is seemingly useless in this scenario, as $1_A$ for $A$ a $\pp$-nullset are zero in $L^\infty$. The characterization of absolute continuity of nonnegative measures $\mu$ through $$\forall\epsilon>0~\exists\delta>0~\forall A\in\scrG~\Big(\pp(A)<\delta\implies\mu(A)<\epsilon\Big)$$ seems more suited for this purpose as a kind of equi-integrability condition on a continuous linear form $L^\infty\to\R$.

Answer 1

There is a straightforward functional analytic argument if you restrict yourself to $L^2(\mathcal{F})$ rather than $L^1(\mathcal{F})$.

If you consider $L^2(\mathcal{G}) \subset L^2(\mathcal{F})$ for the sub sigma algebra $\mathcal{G} \subset \mathcal{F}$ then $Z \mapsto E[ZX]$ is a continuous linear functional on $L^2(\mathcal{F})$ for any $X \in L^2(\mathcal{F})$ where $X$ is the gradient guaranteed by Reisz representation theorem. Now, note that $Z \mapsto E[ZX]$ is also a continuous linear functional on $L^2(\mathcal{G})$. Thus, by Reisz representation theorem, there exists some $E[X|\mathcal{G}] \in L^2(\mathcal{G}) $ such that $E[ZX] = E[Z E[X|\mathcal{G}]]$ for all $Z \in L^2(\mathcal{G})$.

The answer to the following question discusses the $L^1$ to $(L^{\infty})^*$ embedding map and its pseudo-inverse: The Duals of $l^\infty$ and $L^{\infty}$

This might also be relevant: Is there a Banach space version of Riesz representation theorem? https://www.math.ksu.edu/~nagy/real-an/4-06-dual-lp.pdf.