The product $(fg)$ of L-integrable $f,g$ which are measurable on $P$-independent sub-$\sigma$-algebras is integrable

Question

Let $(\Omega, \mathscr{F}, P)$ be a probability space and $\mathscr{A}, \mathscr{G}\subset \mathscr{F}$ be $P$-independant sub-$\sigma$-algebras. Let $f$ be $\mathscr{A}$ measurable and $g$ be $\mathscr{G}$ measurable and $f,g\in \mathscr{L}^1(P)$ and $f,g:\Omega \rightarrow \mathbb{R}$.

I want to show that the product $fg\in \mathscr{L}^1(P)$ and that $$\int_\Omega (fg) dP = (\int_\Omega (f) dP)(\int_\Omega (g) dP)$$ This problem is from my measure theory course, not Stochastics.

For the first part I have the following: $$fg=(f_+-f_-)(g_+-g_-)=f_+g_+-f_+g_--f_-g_++f_-g_-\quad (*)$$ Each term on the r.h.s. is the product of measurable functions, ergo the terms are all measurable. Also, the terms are bounded, as $$f\in \mathscr{L}^1(P)\implies\int_\Omega f_\pm dP < \infty\implies f_\pm < \infty$$ for almost all points in $\Omega$. Together with $P(\Omega)=1$ this is sufficient to show that $fg\in \mathscr{L}^1(P)$

For the second part, my idea was to use $(*)$ to define the integral $\int_\Omega (fg) dP$. At any given point, only one of the terms is non-zero. I thought there may be a way to partition $\Omega$ to use this fact and get a nice expression for the integral.

I'm stuck. The terms of $(*)$ are products, I don't know how to work around that when integrating. Also, some of the terms are negative, and I'm not using the fact that $P(\Omega)=1$, which I think may be important. I only used $P(\Omega)<\infty$.

Is my proof for the first part correct? How can we derive the second part?

Answer 1

$f, g \in \mathscr{L}^1(P)$ doesn't in general imply $fg \in \mathscr{L}^1(P)$, for example$$\int_0^1 x^{-\frac{2}{3}}\textrm{d}x = \left[\frac{1}{3}x^{\frac{1}{3}}\right]_0^1 = \frac{1}{3}\textrm{, but }\int_0^1 \left(x^{-\frac{2}{3}}\right)^2\textrm{d}x = \left[-\frac{1}{3}x^{-\frac{1}{3}}\right]_0^1 = \infty,$$so to show this, you have to make use of the fact that $f$ and $g$ are measurable with respect to mutually independent $\sigma$-algebras.

To show that $\int fg = \int f \cdot \int g$, one can start with simple functions, i.e. measurable functions that take only finitely many distinct values. Let $\phi = \sum_i a_i \mathbf{1}_{A_i}$ be $\mathscr{A}$-measurable, and let $\psi = \sum_j b_j \mathbf{1}_{B_j}$ be $\mathscr{G}$-measurable. Then$$\phi\psi = \sum_{i,j}a_i b_j \mathbf{1}_{A_i \cap B_j}$$and thus$$\int\phi\psi\textrm{d}P = \sum_{i,j}a_i b_j P(A_i \cap B_j) = \sum_i a_i P(A_i) \sum_j b_j P(B_j) = \int\phi\textrm{d}P\int\psi\textrm{d}P.$$

The best way to extend this to arbitrary $f$ and $g$ depends on what facts about integrals you have available. If, for example, you know that non-negative measurable functions can be approximated from below by simple functions and you know the Monotone Convergence Theorem, you can simply write$$\int f_+ \textrm{d}P \int g_+ \textrm{d}P = \lim_{n\to\infty}\int\phi_n\textrm{d}P\cdot\lim_{n\to\infty}\int\psi_n\textrm{d}P = \lim_{n\to\infty}\int\phi_n\psi_n\textrm{d}P = \int f_+g_+ \textrm{d}P$$and conclude that, since the left side is finite, so is the right, and therefore $f_+g_+ \in \mathscr{L}^1(P)$.

The general case for possibly negative $f$ and $g$ then follows from your equation $(*)$, since integration is linear.

Answer 2

It's impossible to prove $fg \in \mathscr L^1(P)$ without using independency: for example, take $\Omega = [0, 1]$ with usual Lebesgue measure, and $f(x) = g(x) = \frac{1}{\sqrt{x}}$. Also, $f_{+}$ is finite, but it doesn't have to be bounded (as in this example).

However, if we can show that for example $f_+ g_+ \in \mathscr L^1(P)$ (and similarly the rest), it will mean that $fg\in \mathscr L^1(P)$, as linear combination of integrable functions is integrable.

To get that $f_+ g_+$ is integrable, and also $\int f_+ g_+\, dP = \left(\int f_+\, dP\right)\left(\int g_+\, dP\right)$, we need to dig a bit into fundamentals of Lebesgue integral. Let $f_+^n$ be some $\mathscr A$-measurable pointwise-increasing sequence of functions that pointwise converges to $f_+$ (for example, we can take $f^n_+(\omega) = k\cdot 2^{-n}$ if $k \cdot 2^{-n} \leq f(\omega) < (k + 1)\cdot 2^{-n}$ for some $k < 4^n$, and $f^n_+(\omega) = 2^n$ if $f(\omega) > 2^n$), and similarly for $g_+^n$.

By independence, $$\int f_+^n g_+^n\, dP = \left(\int f_+^n\,dP\right)\left(\int g_+^n\,dP\right)$$, and $f_+^n g_+^n \to f_+g_+$ pointwise and monotonically. By Beppo Levi's monotone convergence theorem, we get $f_+ g_+$ is measurable, and

$$\lim_{n\to \infty}\int f_+^n g_+^n\, dP = \int f_+g_+\, dP$$ $$\lim_{n \to \infty}\int f_+^n\,dP = \int f_+\, dP$$ $$\lim_{n \to \infty}\int g_+^n\,dP = \int g_+\, dP$$

Combining these 4 formulas and using limit of product we get $$\int f_+g_+\, dP = \left(\int f_+\,dP\right)\left(\int g_+\,dP\right)$$

Also, $P(\Omega) = 1$ is used when we introduce independent algebras. As $\Omega \in \mathscr A, G$, we necessary have $P(\Omega)\cdot P(\Omega) = P(\Omega)$, thus $P(\Omega) \in \{0, 1\}$, and case $P(\Omega) = 0$ isn't interesting.