$E(XY)=E(X)E(Y)$ and pull out property of conditional expectation without standard machine

Question

This is gonna sound dumb but here goes.

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From David Williams' Probability with Martingales:

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Incorrect proof:

$$\int_{\Omega} X d \mathbb P \int_{\Omega} Y d \mathbb P$$

$$ = \int_{\Omega} \underbrace{\int_{\Omega} XY d \mathbb P}_{a \ number} d \mathbb P$$

$$ = \int_{\Omega} XY d \mathbb P \int_{\Omega} (1) d \mathbb P$$

$$ = \int_{\Omega} XY d \mathbb P (1)$$

$$ = \int_{\Omega} XY d \mathbb P $$

So what's wrong here? I guess:

1. The first step is nonsensical.
2. The first step is not nonsensical but wrong.
3. The first step is not nonsensical, right but at that point in the text, we don't know that we can do that. It may be explained in Chapter 8, but I'm not sure I can find exactly.

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I think I can similarly prove $E[YX | \mathscr G] = YE[X|\mathscr G]$ without standard machine:

$E[X|\mathscr G]$ is any $\mathscr G$-measurable random variable $H$ s.t. $$\int_G H d \mathbb P =\int_G X d \mathbb P \tag{*}$$

$E[XY|\mathscr G]$ is any $\mathscr G$-measurable random variable $Z$ s.t. $$\int_G Z d \mathbb P =\int_G XY d \mathbb P$$

From $(*)$, I multiply $\int_G Y d \mathbb P$ on both sides:

$$\int_G Y d \mathbb P \int_G H d \mathbb P =\int_G Y d \mathbb P \int_G X d \mathbb P$$

$$\int_G \int_G H Y d \mathbb P d \mathbb P =\int_G \int_G X Y d \mathbb P d \mathbb P$$

$$\int_G H Y d \mathbb P = \int_G X Y d \mathbb P $$

$$\int_G H Y d \mathbb P = \int_G Z d \mathbb P $$

QED?

Okay now I sense something wrong here like if we could just manipulate Lebesgue integrals this way, then everything would be equal or something. What exactly is wrong please?

Answer 1

The error is calling that function a number. Use dummy variables. $$\int_\Omega X(\omega)d\mathbb{P}(\omega) \int_\Omega Y(\nu)d\mathbb{P}(\nu) = \int_\Omega \int_\Omega X(\omega)Y(\nu)d\mathbb{P}(\omega)d\mathbb{P}(\nu).$$ The term $$\int_\Omega X(\omega)Y(\nu)d\mathbb{P}(\omega)$$ is not a number but rather a function of $\nu$.