Show that this is a random inner product

Question

Fix a probability space $(\Omega,\mathcal A,P)$ and let $\mathcal F $ be a sub $\sigma$-algebra of $\mathcal A$. Let $\mathcal P^+$ denote the set of all random variables $X$ on $(\Omega,\mathcal A,P)$ satisfying $E[X^2\mathcal |\mathcal F]<\infty$ a.s., where $E[X^2\mathcal |\mathcal F]$ is a nonnegative extended real-valued conditional expectation.

From the conditional Minkowski's inequality and the pull-out property of conditional expectations, we have that $Z_1X_1+Z_2X_2\in \mathcal P^+$ whenever $X_1,X_2\in P^+$ and $Z_1,Z_2$ are $\mathcal F$-measurable.

For $X,Y$ in $\mathcal P^+$ define $\langle X,Y\rangle:=E[(XY)^+\mathcal |\mathcal F]-E[(XY)^-\mathcal |\mathcal F]$, where $E[(XY)^+\mathcal |\mathcal F]$, $E[(XY)^-\mathcal |\mathcal F]$ are chosen to be nonnegative real-valued conditional expectations. This is possible because, by the conditional Hölder's inequality, we have $E[|XY|\mathcal |\mathcal F]\leq E[X^2\mathcal |\mathcal F]^{1/2} E[Y^2\mathcal |\mathcal F]^{1/2}<\infty$ a.s.. For $X$ in $\mathcal P^+$ also define $\|X\|:=E[X^2\mathcal |\mathcal F]^{1/2}$ with $E[X^2\mathcal |\mathcal F]$ is chosen nonnegative and real-valued. Note that $\langle X,Y\rangle$ and $\|X\|$ are only defined a.s..

Now am asked to show the following properties for $X,Y,Z$ in $\mathcal P^+$ :

1. $\langle X,Y\rangle=\langle Y,X\rangle$ a.s..

2. $\langle X+Y,Z\rangle=\langle X,Z\rangle+\langle Y,Z\rangle$ a.s..

3. $\langle ZX,Y\rangle=Z\langle Y,X\rangle$ a.s. if $Z$ is $\mathcal F$-measurable.

4. $\langle X,X\rangle\geq 0$ a.s. and $\langle X,X\rangle= 0$ a.s. if and only if $X=0$ a.s..

5. $\|X\|=\sqrt{\langle X,X\rangle}$ a.s..

6. $|\langle X,Y\rangle|\leq \|X\|\|Y\|$ a.s..

7. $\|X+Y\|\leq \|X\|+\|Y\|$ a.s.

8. $\|ZX\|=|Z|\|X\|$ a.s. if $Z$ is $\mathcal F$-measurable.

9. $\|X\|\geq 0$ and $\|X\|=0$ a.s. if and only if $X=0$ a.s..

My attempt:

1. Clear.

2. We have the identity $$((X+Y)Z)^++(XZ)^-+(YZ)^-=((X+Y)Z)^-+(XZ)^++(YZ)^+$$ Taking conditional expectations on both sides and using linearity we get $$E[((X+Y)Z)^+|\mathcal F]+E[(XZ)^-|\mathcal F]+E[(YZ)^-|\mathcal F]=E[((X+Y)Z)^-|\mathcal F]+E[(XZ)^+|\mathcal F]+E[(YZ)^+|\mathcal F] \quad \text{a.s.}$$ Since everything is finite a.s. we get the result by subtracting.

3. First suppose $Z\geq 0$. Then $E[(ZXY)^+\mathcal |\mathcal F]=ZE[(XY)^+\mathcal |\mathcal F]$ and $E[(ZXY)^-\mathcal |\mathcal F]=ZE[(XY)^-\mathcal |\mathcal F]$ a.s. by the pull-out property. Taking difference we get $\langle ZX,Y\rangle=Z\langle Y,X\rangle$ a.s.. Similarly if $Z\leq 0$ then $\langle ZX,Y\rangle=Z\langle Y,X\rangle$ a.s.. Finally for arbitrary $Z$ we have $\langle ZX,Y\rangle=\langle Z^+X,Y\rangle+\langle -Z^-X,Y\rangle=Z^+\langle X,Y\rangle-Z^-\langle X,Y\rangle=Z\langle X,Y\rangle$ a.s. using 2. and the previous results.

4. $\langle X,X\rangle=E[X^2|\mathcal F]\geq 0$ a.s. and $E[X^2|\mathcal F]=0$ a.s. if and only if $X=0$ a.s..

5. $\sqrt{\langle X,X\rangle}=E[X^2|\mathcal F]^{1/2}=\|X\|$ a.s..

6. $|\langle X,Y\rangle|\leq E[(XY)^+\mathcal |\mathcal F]+E[(XY)^-\mathcal |\mathcal F]=E[|XY|\mathcal |\mathcal F]\leq E[X^2\mathcal |\mathcal F]^{1/2} E[Y^2\mathcal |\mathcal F]^{1/2}=\|X\|\|Y\|$ a.s. by the conditional Hölder's inequality.

7. $\|X+Y\|=E[(X+Y)^2|\mathcal F]^{1/2}\leq E[X^2\mathcal |\mathcal F]^{1/2}+ E[Y^2\mathcal |\mathcal F]^{1/2}=\|X\|+ \|Y\|$ a.s. by the conditional Minkowski's inequality.

8. $\|X\|=\sqrt{\langle ZX,ZX\rangle}=\sqrt{Z^2\langle X,X\rangle}=|Z|\|X\|$ a.s. by 5,3,1.

9. Follows from 5 and 4.

Am I missing something?