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If $\langle \Omega, \mathfrak{F}, \mathbb{P}\rangle$ is a measure space and $L^2$ is the corresponding $L^2$ space and $$ K\oplus K^{\perp} \cong L^2(\mathfrak{F},\mathbb{P}). $$

Then let: $$ \mathfrak{F}_1\triangleq \sigma(\{H \in K \}) \\ \mathfrak{F}_2\triangleq \sigma(\{H \in K^{\perp} \}) \\ $$

Moreover, do there exist singular measures $\mu$ and $\nu$ such that $$ \mu + \nu = \mathbb{P},\\ L^2(\mathfrak{F}_1,\mu) \oplus L^2(\mathfrak{F}_2,\nu) \cong L^2(\mathfrak{F},\mathbb{P}) $$

$$ \int f d\nu =0 = \int g d\mu $$ if $f \in L^2(\mathfrak{F}_1)$ and $g \in \mathfrak{F}_2$?

I was thinking of using conditional measures, but I'm not sure how?

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  • $\begingroup$ For a counterexample to the first question, can't you simply take your favorite pair of random variables $f,g$ which are mean-zero and uncorrelated but not independent, and let $K$ be the one-dimensional space spanned by $f$? $\endgroup$
    – Nate Eldredge
    Commented Nov 4, 2016 at 1:58
  • $\begingroup$ Alternatively, let $(\Omega, \mathcal{F}, \mathbb{P})$ be $[0,1]$ with the Borel $\sigma$-field and Lebesgue measure, and let $K$ be the one-dimensional space spanned by $f(x) = x$, or your other favorite one-to-one function. It should be easy to show that $\mathcal{F}_1 = \mathcal{F}_2 = \mathcal{F}$. $\endgroup$
    – Nate Eldredge
    Commented Nov 4, 2016 at 2:03
  • $\begingroup$ Ok i removed that, how about the second part? $\endgroup$
    – Rombaldo Meniscus
    Commented Nov 4, 2016 at 2:08
  • $\begingroup$ What does $\cong$ mean? Isomorphic? Isometric? Isometric via the natural map? $\endgroup$
    – Nate Eldredge
    Commented Nov 4, 2016 at 2:08
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    $\begingroup$ No, they will not be orthogonal in general. See the example above, in which they are equal. $\endgroup$
    – Nate Eldredge
    Commented Nov 4, 2016 at 2:44

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I'm not sure I have completely parsed your question, but it seems to be based on an assumption that the $\sigma$-fields $\mathfrak{F}_1, \mathfrak{F}_2$ are in some sense "orthogonal". That doesn't have to be true; they can even be equal.

Take as an example $\Omega = [0,1]$ with its Borel $\sigma$-field $\mathcal{F}$ and let $\mathbb{P}$ be Lebesgue measure. Note the following fact: if $g : [0,1] \to \mathbb{R}$ is Borel and injective with a Borel left inverse $g^{-1} : \mathbb{R} \to [0,1]$, then $\sigma(g)$, the smallest $\sigma$-field making $g$ measurable, is $\mathfrak{F}$ itself. (Proof: for any Borel set $B \subset [0,1]$, the function $1_B \circ g^{-1} \circ g = 1_B$ is $\sigma(g)$-measurable.)

Let $g_1(x) = x$ and $g_2(x) = x^2 - \frac{1}{2}$. They are orthogonal in $L^2$, so if we let $K$ be the one-dimensional space spanned by $g_1$, we have $g_2 \in K^\perp$. But each is Borel and injective (indeed, each is a homeomorphism onto its image) so we end up with $\mathfrak{F}_1 = \mathfrak{F}_2 = \mathfrak{F}$.

In particular, if $\nu$ is a measure absolutely continuous to $\mathbb{P}$, and $\int f\,d\nu = 0$ for all $f \in L^2(\mathfrak{F}_1)$, then $\nu = 0$.

Basically, the issue is that there are a lot of operations that don't enlarge a generated $\sigma$-field, but do enlarge a subspace of $L^2$. Multiplication is perhaps the most obvious example.

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