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Your desired conclusion is correct, provided that the functions $f$ and $g$ are assumed to be Borel measurable.

Indeed, let $X:=f(\bf x)-g(\bf x)$ and $Z:=\bf z$, so that $X$ and $Z$ are bounded random variables with values in $\mathbb R$ and $\mathbb R^n$, respectively, such that $$EXP(Z)=0$$ for all polynomials $P$ on $\mathbb R^n$. Since the set of all such polynomials is dense in $L^1(K)$ for any compact $K\subset\mathbb R^n$, whereas $X$ and $Z$ are bounded, we conclude that $EXh(Z)=0$ for any $h\in L^1(K)$. So, $EX\,1(Z\in B)=0$ for any Borel set $B\subseteq\mathbb R^n$, which means $E(X|Z)=0$, as desired.

Your desired conclusion does hold without the assumption of a probability density of $\bf x$ and $\bf z$, provided that the functions $f$ and $g$ are assumed to be Borel measurable.

Indeed, let $X:=f(\bf x)-g(\bf x)$ and $Z:=\bf z$, so that $X$ and $Z$ are bounded random variables with values in $\mathbb R$ and $\mathbb R^n$, respectively, such that $$EXP(Z)=0$$ for all polynomials $P$ on $\mathbb R^n$. Since the set of all such polynomials is dense in $L^1(K)$ for any compact $K\subset\mathbb R^n$, whereas $X$ and $Z$ are bounded, we conclude that $EXh(Z)=0$ for any $h\in L^1(K)$. So, $EX\,1(Z\in B)=0$ for any Borel set $B\subseteq\mathbb R^n$, which means $E(X|Z)=0$, as desired.

It is unclear how you define the expressions $E\left[f({\bf x})\delta({\bf z} -{\bf z}_0)\right]$ and $E\left[f({\bf x})|{\bf z}_0\right]$. However, your desired conclusion is correct, provided that the functions $f$ and $g$ are assumed to be Borel measurable.

Indeed, let $X:=f(\bf x)-g(\bf x)$ and $Z:=\bf z$, so that $X$ and $Z$ are bounded random variables with values in $\mathbb R$ and $\mathbb R^n$, respectively, such that $$EXP(Z)=0$$ for all polynomials $P$ on $\mathbb R^n$. Since the set of all such polynomials is dense in $L^1(K)$ for any compact $K\subset\mathbb R^n$, whereas $X$ and $Z$ are bounded, we conclude that $EXh(Z)=0$ for any $h\in L^1(K)$. So, $EX\,1(Z\in B)=0$ for any Borel set $B\subseteq\mathbb R^n$, which means $E(X|Z)=0$, as desired.

Your desired conclusion is correct, provided that the functions $f$ and $g$ are assumed to be Borel measurable.

Indeed, let $X:=f(\bf x)-g(\bf x)$ and $Z:=\bf z$, so that $X$ and $Z$ are bounded random variables with values in $\mathbb R$ and $\mathbb R^n$, respectively, such that $$EXP(Z)=0$$ for all polynomials $P$ on $\mathbb R^n$. Since the set of all such polynomials is dense in $L^1(K)$ for any compact $K\subset\mathbb R^n$, whereas $X$ and $Z$ are bounded, we conclude that $EXh(Z)=0$ for any $h\in L^1(K)$. So, $EX\,1(Z\in B)=0$ for any Borel set $B\subseteq\mathbb R^n$, which means $E(X|Z)=0$, as desired.

It is unclear how you define the expressions $E\left[f({\bf x})\delta({\bf z} -{\bf z}_0)\right]$ and $E\left[f({\bf x})|{\bf z}_0\right]$. However, your desired conclusion is correct, provided that the functions $f$ and $g$ are assumed to be Borel measurable.

Indeed, let $X:=f(\bf x)-g(\bf x)$ and $Z:=\bf z$, so that $X$ and $Z$ are bounded random variables in $\mathbb R$ and $\mathbb R^n$, respectively, such that $$EXP(Z)=0$$ for all polynomials $P$ on $\mathbb R^n$. Since the set of all such polynomials is dense in $L^1(K)$ for any compact $K\subset\mathbb R^n$, whereas $X$ and $Z$ are bounded, we conclude that $EXh(Z)=0$ for any $h\in L^1(K)$. So, $EX\,1(Z\in B)=0$ for any Borel set $B\subseteq\mathbb R^n$, which means $E(X|Z)=0$, as desired.

It is unclear how you define the expressions $E\left[f({\bf x})\delta({\bf z} -{\bf z}_0)\right]$ and $E\left[f({\bf x})|{\bf z}_0\right]$. However, your desired conclusion is correct, provided that the functions $f$ and $g$ are assumed to be Borel measurable.

Indeed, let $X:=f(\bf x)-g(\bf x)$ and $Z:=\bf z$, so that $X$ and $Z$ are bounded random variables with values in $\mathbb R$ and $\mathbb R^n$, respectively, such that $$EXP(Z)=0$$ for all polynomials $P$ on $\mathbb R^n$. Since the set of all such polynomials is dense in $L^1(K)$ for any compact $K\subset\mathbb R^n$, whereas $X$ and $Z$ are bounded, we conclude that $EXh(Z)=0$ for any $h\in L^1(K)$. So, $EX\,1(Z\in B)=0$ for any Borel set $B\subseteq\mathbb R^n$, which means $E(X|Z)=0$, as desired.