Exercise 1.9: Show that if a random variable X is independent of an event A, then a function of the random variable g(X) is also independent of A

Question

I am undertaking a self study of Stochastic Calculus for Finance II by Shreve. I have run into a problem I can't figure out how to solve. For reference, the full text can be found at the following link: https://cms.dm.uba.ar/academico/materias/2docuat2016/analisis_cuantitativo_en_finanzas/Steve_ShreveStochastic_Calculus_for_Finance_II.pdf

The problem is as follows:
Exercise 1.9. Suppose X is a random variable on some probability space ($\Omega$, F, $\mathbb{P}$), A is a set in F, and for every Borel subset B of $\mathbb{R}$, we have
$$\int_A \mathbb{I}_B(X(\omega))d\mathbb{P}(\omega) = \mathbb{P}(A)*\mathbb{P}\{X\in B\}$$
Then we say that X is independent of the event A.

Show that if X is independent of an event A, then
$$\int_A g(X(\omega))d\mathbb{P}(\omega) = \mathbb{P}(A)*\mathbb{E}g(X)$$ for every nonnegative, Borel-measurable function g

Note: I know the following theorem is supposed to be used in proving this statement. (It can be found on page 28 (slide 48) and the proof of the theorem follows on pages 28-30 (slides 48-50) of the book (link I provided)).

Theorem 1.5.1 Let X be a random variable on a probability space ($\Omega$, F, $\mathbb{P}$) and let g be a Borel-measurable function on $\mathbb{R}$. Then
$$\mathbb{E}|g(X)| = \int_\mathbb{R}|g(X)|d\mu_X(x)$$
and if this quantity is finite, then
$$\mathbb{E}g(X) = \int_\mathbb{R}g(X)d\mu_X(x)$$

Any help would be greatly appreciated. Also, in case anyone was wondering... I am aware of the solutions manual by Yan Zeng (I'll post a link to it at the end of this incase anyone wants to take a look at it). Unfortunately, it doesn't seem to be much help on this problem. Just for clarification, after going through the section I feel as though I understand why the statement is true, which is why the solution manual isn't a lot of help in this case, because it just gives an outline for the process for proving it. I'm stuck on the process of how to exactly show that the statement is true.

Solutions manual can be found at the following link: https://www.quantsummaries.com/shreve_stochcal4fin_2.pdf ____________________________________________________________ ____________________________________________________________ Edit: Here is what I have as a solution so far, I'm just not sure about one step (which I will point out when it comes up)

Case 1: Suppose $g(x) = \mathbb{I}_B(x)$
Then $\int_A g(X(\omega))d\mathbb{P}(\omega) = \int_A \mathbb{I}_B(X(\omega))d\mathbb{P}(\omega) = \mathbb{P}(A)*\mathbb{P}\{X\in B\}$ (by def. of independence and assumption)

Case 2: Suppose g(x) is a nonnegative simple function $$g(x) = \sum_{k=1}^n \alpha_k\mathbb{I}_{B_k}(x)$$ where $\alpha_1, \alpha_2,...,\alpha_n$ are nonnegative constants & $B_1,..., B_n$ are Borel subsets of $\mathbb{R}$

Then:

$\int_A g(X(\omega))d\mathbb{P}(\omega) = \int_A\sum_{k=1}^n\alpha_k\mathbb{I}_{B_k}(x)d\mu_X(x) = \sum_{k=1}^n\alpha_k\int_A\mathbb{I}_{B_k}(x)d\mu_X(x)$

$ = \sum_{k=1}^n\alpha_k\mathbb{P}\{X\in B\}\mathbb{P}(A) = \mathbb{P}(A)\sum_{k=1}^n\alpha_k\mathbb{P}\{X\in B\} = \mathbb{P}(A)\mathbb{E}g(X) $

NOTE: The last step is the part I am still confused on. If I've made a mistake, I'm not sure where. If I'm correct, then I just can't seem to wrap my head around why the last step is true. As far as the rest of the proof, there's only one more case to prove (since I'm only proving the statement is true for nonnegative functions g) and it follows directly from Case 2 and the Monotone Convergence Theorem.

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P.S. This is my first post on Stack Exchange so please feel free to give me feedback on my question and let me know if there's more information I should provide or anything like that/ general etiquette for posting questions.

Answer 1

$\int_A g(X(\omega))d\mathbb{P}(\omega) = \int_A\sum_{k=1}^n\alpha_k\mathbb{I}_{B_k}(x)d\mu_X(x) = \sum_{k=1}^n\alpha_k\int_A\mathbb{I}_{B_k}(x)d\mu_X(x)$ $ = \sum_{k=1}^n\alpha_k\mathbb{P}\{X\in \color{red}{B}\}\mathbb{P}(A) = \mathbb{P}(A)\sum_{k=1}^n\alpha_k\mathbb{P}\{X\in \color{red}{B}\} = \mathbb{P}(A)\mathbb{E}g(X) $

The mistakes are highlighted in red. Each $B$ should be $B_k$. Also,

$$\Bbb{P}(A)\sum_{k = 1}^n \alpha_k\Bbb{P}\{X\in B_k\} = \Bbb{P}(A)\sum_{k = 1}^n \alpha_k \Bbb{E}(\Bbb{I}_{B_k}\circ X) = \Bbb{P}(A)\,\Bbb{E}\left[\sum_{k = 1}^n \alpha_k \Bbb{I}_{B_k}\circ X\right] = \Bbb{P}(A)\Bbb{E}g(X)$$