Fubini's theorem problem 
Let $f$ be a non-negative measurable function in $\mathbb{R}$. Suppose that $$\iint_{\mathbb{R}^2}{f(4x)f(x-3y)\,dxdy}=2\,.$$ Calculate $\int_{-\infty}^{\infty}{f(x)\,dx}$.

My first thought was to change the order of integration so that I integrate in $y$ first, since there's only a single $y$ in the integrand... but I'm not sure how/if that even helps me. 
Then the more I thought about it, the less clear it was to me that Fubini's theorem even applies as it's written. Somehow I need a function of two variables. So should I set $g(x,y) = f(4x)f(x-3y)$ and do something with that? At least Fubini's theorem applies for $g(x,y)$, since we know it's integrable on $\mathbb{R}^2$.    .... Maybe?
I'm just pretty lost on this, so any help you could offer would be great. Thanks!
 A: Your thought is a good one.  If you integrate in $y$ first you can change variable to $u=x-3y$.  You are considering $x$ a constant, so you get a decoupled product of integrals of $f$ over the real line.
A: I think both Vitali and Matt are right. As soon as $G(x,y)$ is integrable on $\mathbb{R}^2$, $$\iint_{\mathbb{R}^2}{G(x,y)\,dxdy}=\int_{\mathbb{R}}dx\,\int_{\mathbb{R}}{G(x,y)\,dy}=\int_{\mathbb{R}}dy\,\int_{\mathbb{R}}{G(x,y)\,dx} $$
So you can substitute: $$ u=4x $$ $$ v=x-3y $$ $$ x=\frac{u}{4} $$ $$ y=\frac{u}{12}-\frac{v}{3} $$
Jacobian $$ J=\det D(x(u,v),y(u,v))=\begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}=\begin{vmatrix}\frac{1}{4} & 0\\\\\frac{1}{12} & -\frac{1}{3}\end{vmatrix}=-\frac{1}{12} $$
Then $$\iint_{\mathbb{R}^2}{f(4x)\,f(x-3y)\,dxdy}=-\frac{1}{12}\int_{-\infty}^{\infty}\,du\,\int_{\infty}^{-\infty}{f(u)\,f(v)\,dv}=\frac{1}{12}(\int_{-\infty}^{\infty}{f(z)\,dz})^2=2$$
Thus giving us
$$\int_{-\infty}^{\infty}{f(z)\,dz}=2\sqrt 6$$
Please tell me if I $\mathbb{F}\bigoplus$ something up.
