Csdrhrt

If $X_1, ..., X_{n+1}$ are independent real random variables and $h:mathbb{R}^{n+1} to mathbb{R}$ a Borel function. Now taking the conditional expectation $mathbb{E}[h(X_1, ldots, X_{n+1})| sigma(X_1, ldots, X_n)]$

Assuming $mathbb{E}[|h(X_1, ldots, X_{n+1})|] < infty $ show that if
$g(x_1, ldots, x_n) = mathbb{E}[h(x_1, ldots, x_n, X_{n+1})]$ then
$g(X_1, ldots, X_n)$ is a version of $mathbb{E}[h(X_1, ldots, X_{n+1})| sigma(X_1, ldots, X_n)]$.

The claim would follow from the definition of conditional expectation, if we can show that
$$ mathbb{E}[h(X_1, ldots, X_{n+1}) mathbf{1}_G] = mathbb{E}[g(X_1, ldots, X_{n}) mathbf{1}_G], $$
for all $G in sigma(X_1, ldots, X_n)$.

What I tried so far is trying to expand the expectations and rewrite them with Fubini's theorem.

Assuming the underlying probability space is $(Omega, mathcal{F}, P)$.
I thought I could rewrite with a restricted push forward $P_{|G}^{X_i}(A) = P(X_i^{-1}(A) cap G)$ to get
$$ mathbb{E}[h(X_1, ldots, X_{n+1}) mathbf{1}_G] = int_{(x_1, ldots x_{n+1})in mathbb{R}^{n+1}} h(x_1, ldots, x_{n+1}) dP_{|G}^{X_1, ldots, X_{n+1}} $$
All variables are $X_i$ independet so we can factor $dP_{|G}^{X_1, ldots, X_{n+1}} = d Pi_{i=1}^{n+1} P_{|G}^{X_i}$.

$$ = int_{(x_1, ldots x_{n+1})in mathbb{R}^{n+1}} h(x_1, ldots, x_{n+1}) d Pi_{i=1}^{n+1} P_{|G}^{X_i} = int_{(x_1, ldots x_{n})in mathbb{R}^{n}} left(int_{x_{n+1}in mathbb{R}} h(x_1, ldots, x_{n+1}) dP_{|G}^{X_{n+1}} right) d Pi_{i=1}^{n} P_{|G}^{X_i} $$

Now $P_{|G}^{X_{n+1}}(A) = P(X_{n+1}^{-1}(A)cap G) = P(X_{n+1}^{-1}(A))P(G)$, since $sigma(X_{n+1})$ and $sigma(X_1,ldots, X_n)$ are independet. Since $mathbb{E}[|h(X_1, ldots, X_{n+1})|] < infty $ Fubini applies and we get

$$ int_{(x_1, ldots x_{n})in mathbb{R}^{n}} left(int_{x_{n+1} in mathbb{R}} h(x_1, ldots, x_{n+1}) dP_{|G}^{X_{n+1}} right) d Pi_{i=0}^{n} P_{|G}^{X_i} = P(G)int_{(x_1, ldots x_{n})in mathbb{R}} left(int_{x_{n+1} in mathbb{R}} h(x_1, ldots, x_{n+1}) dP^{X_{n+1}} right) d Pi_{i=0}^{n} P_{|G}^{X_i} $$
$$ = P(G) int_{(x_1, ldots x_{n}) in mathbb{R}^n} E[h(x_1, ldots, X_{n+1})] d Pi_{i=0}^{n} P_{|G}^{X_i} = P(G) int_{(x_1, ldots x_{n})in mathbb{R}^n} g(x_1, ldots, x_n) d Pi_{i=0}^{n} P_{|G}^{X_i} = P(G)mathbb{E}[g(X_1, ldots, X_n) mathbf{1}_G]. $$

So I the end I get
$$ mathbb{E}[h(X_1, ldots, X_{n+1}) mathbf{1}_G] = P(G)mathbb{E}[g(X_1, ldots, X_n) mathbf{1}_G],$$ which is wrong according to the claim!

I can't spot my mistake. So is the claim really true and if so where do I mess up?

Denote $X=(X_1,dots,X_n)sim mu$ and $Y=X_{n+1}sim nu$, then $X,Y$ are independent and $(X,Y)sim mu times nu$. Also,
$$ g(x)=int_{mathbb{R}}h(x,y)nu(dy) quad xin mathbb{R}^n.$$ Thus $$g(X)=mathbb{E}[h(X,Y)|X].$$

In fact, $g(X)$ is $sigma(X)$-measurable and
$$ mathbb{E}[g(X)1_{{Xin A}}]=
mathbb{E}[g(X)1_A(X)]= int_{mathbb{R}^n} g(x)1_A(x) mu(dx)=int_{mathbb{R}^n} left( int_{mathbb{R}} h(x,y)nu(dy)right)1_A(x)mu(dx) = int_{mathbb{R}^ntimes mathbb{R}} h(x,y)1_A(x)mutimes nu(dxtimes dy) = mathbb{E}[h(X,Y)1_A(X)]=
mathbb{E}[h(X,Y)1_{{Xin A}}]$$ for all $Ain mathcal{B}_{mathbb{R}^n}$.