Suppose X and Y are independent.Let $\varphi $ be a function with $E(|\varphi(X,Y)|)< \infty$ and let $g(x)=E(\varphi(x,Y))$.The conclusion is $E(\varphi(X,Y)|X)=g(X)$
So the first step is to show $g(X)\in \sigma(X)$ and I don't know how to do.
There is a theorem "If $g$ is a continuous function ,then $g(X)\in \sigma(X)$".It seems no other theorems about measurable.But I think $g$ need not to be continuous.
$g$ need not be continuous, but it has to be a Borel-measurable function $g: (\mathbb{R},\mathcal{B}(\mathbb{R})) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$. In order to conclude that $g(X)$ is $\sigma(X)/\mathcal{B}(\mathbb{R})$-measurable, note that $$(\Omega,\sigma(X)) \ni \omega \mapsto X(\omega) \in (\mathbb{R},\mathcal{B}(\mathbb{R}))$$ is measurable and that the composition of two measurable functions is measurable (with respect to the corresponding $\sigma$-algebras).