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Context: these lecture notes, exercise 46. Not a homework assignment.

The notes give the following hint:

If $f,g$ are independent RVs, then $\mathbb{E}[fg] = \mathbb{E}[f]\mathbb{E}[g]$.

Here's my attempt:

for some $A \in \mathcal{G}$, take $g = \mathbb{1}_A$. As $f$ is independent of $\mathcal{G}$, $f$ is independent of $\mathbb{1}_A$. So $$\mathbb{E}[f\mathbb{1}_A] = \mathbb{E}[f]\mathbb{E}[\mathbb{1}_A]= \mathbb{E}[f]\mu(A)$$ $$\frac{\mathbb{E}[f\mathbb{1}_A]}{\mu(A)} = \mathbb{E}[f]$$ $$\mathbb{E}[f|A] = \mathbb{E}[f].$$

This holds for any $A\in \mathcal{G}$, but I can't see how it implies the thesis.

Alternatively, I was thinking of using the indicator function on the sigma algebra $\mathcal{G}$ as my $g$ in the formula, although does "the indicator function on the sigma algebra" actually mean anything?

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To show $E[f \mid \mathcal{G} ] = E[f]$, just show that $E[f]$ satisfies the necessary conditions of the conditional expectation $E[f \mid \mathcal{G}]$. Equality will follow by uniqueness of conditional expectations.

Okay, so for $E[f]$ to be $E[f \mid \mathcal{G}]$, we would need that:

  1. $E[f]$ is $\mathcal{G}$-measurable.
  2. For each $A \in \mathcal{G}$, $\int_{A} E[f] \,d\mu = \int_{A} f \,d\mu$.

Condition 1 is satisfied since $E[f]$ is just a number/constant.

For condition 2, let $A \in \mathcal{G}$. We use the hint similar to how you used it:

$$\int_{A} E[f] \,d\mu = E[f] \int_{A} \,d\mu = E[f] E[\Bbb 1_{A}] = E[f \Bbb 1_{A}] = \int_{A} f \,d\mu$$

and this shows condition 2 is satisfied. Note that the first equality in the above string of equalities is obtained by recognizing that $E[f]$ is just a constant number, so we can pull it out of the integral.