Can dx=dy happen when x and y are independent? My question is pretty simple. can $dx=dy$ happen when x and y are independent random variable? 
What I want to do here is to show that $\int f(x,y) dy =\int f(x,y) dx$, which are the marginal probability density function. I have assumed that x and y are independent random variables. 
I guess this $\int f(x,y) dy =\int f(x,y) dx$ only satisfies when $dx=dy$.
 A: The term $\int f(x,y)\,dx$ is a function of $y$ - let's call if F(y) and we compute this function as
$$F(y) = \int_{-\infty}^{\infty} f(x,y)\,dx$$
whereas the term $\int f(x,y)\,dy$ is a function of $x$ - let's call if G(x) and we compute this function as
$$G(x) = \int_{-\infty}^{\infty} f(x,y)\,dy$$
It is possible that $F$ and $G$ are of identical function form (e.g, if $f(x,y)=f(x+y)$).  However, the notation $\int f dx$ and $\int f dy$ are actually limits of Riemann sums.  
In the former, the partition is taken along the x-axis while in the latter, the partition is taken along the y-axis.  These partitions are not related to one another and thus, it is meaningless to talk about their equality.

NOTE.
The density function for the case of independent random variables may be written as $f(x,y)=f_x(x)f_y(y)$.  Thus, the marginal density functions $F$ and $G$ are 
$$F(y)=\int f(x,y)dx=\int f_x(x)f_y(y)dx=f_y(y)\int f_x(x)dx$$
$$G(x)=\int f(x,y)dy=\int f_x(x)f_y(y)dy=f_x(x)\int f_y(y)dy$$
