Given a filtered probability space $(\Omega, \mathcal{F}, (\mathcal{F}_t)_t, P)$ and with notation (regular conditional probabiltiy assumed): $$P^{X|\mathcal{F}_t}(B)=P(X\in B|\mathcal{F}_t)=E(1_B(X)|\mathcal{F}_t),$$ how can I prove that for a real-valued measurable function $f$ and a $\mathcal{F}_t$-measurable random variable $Y$ we have:$$E(f(X,Y)|\mathcal{F}_t)=\int f(x,Y)P^{X|\mathcal{F}_t}(dx)?$$
My attempt
I think one could approach by first showing it for indicators and than working one's way up. But what would the indicator in this case look like (in the $2$-dim case)?
$$E[f(X,Y)|\mathcal{F}_t]=\int f(X,Y)dP_{\Omega /\mathcal{F}_t}=\int f(x,y)dP_{X\circ Y/\mathcal{F}_t}=\int f(x,Y)dP_{X/\mathcal{F}_t}$$ Since, being $Y$ $\mathcal{F}_t$-measurable: $$P_{X\circ Y/\mathcal{F}_t}(A\times B)=I_B(Y)P_{X/\mathcal{F}_t}(A)$$