What is wrong with below proof that E[XY|X] not equal to E[Y|X]

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The equality above is wrong. But what was the issue with below derivation? Set $T=XY$ $$ E[YX|X=x]=\int_a^b t f_{T|X=x}(t) dt $$ perform change of variables, $dt=x dy$. So
$$ E[YX|X=x]=\int_{\frac{a}{x}}^{\frac{b}{x}} yx f_{T|X=x}(yx) x dy. $$ Now, so $$ f_{T|X=x}(yx)=\frac{1}{x}f_{Y|X=x}(y). $$ Therefore $$ E[YX|X=x]=\int_{\frac{a}{x}}^{\frac{b}{x}} yx f_{Y|X=x}(y) dy=x\int_{\frac{a}{x}}^{\frac{b}{x}} y f_{Y|X=x}(y) dy\ne x E[Y|X=x]. $$ Thanks to Sangchul and Henry, some issues were fixed.


Using joint density: \begin{align*} \int_{R(X)}\int_{R(T)} t \frac{f_{T,X}(t,x)}{f_X(x)}dt dx&=\int_{R(X)}\int_{R(T)/x} xy \frac{f_{XY,X}(xy,x)}{f_X(x)}|x|dy dx \end{align*} Similar problem. So it's the integration bounds...


Sorry, if range of $XY$ is on $a$ to $b$... range of $Y$ changes with $X$.