Question: If $1>1-ab>0$, which of the following must be true?

1. $\frac ab >0$
2. $\frac ab <1$
3. $ab <1$

Option 1 and 3 are correct, but is option 2 necessarily correct?

Answer: 
By the Fubini's theorem, we can change the order of integration. We have that 

$\int\_{\mathbb{R}}\int\_{\mathbb{R}}f(x-t)g(t)dtdx=\int\_{\mathbb{R}}\int\_{\mathbb{R}}f(x-t)g(t)dxdt$

Now, let $u=x-t$. Then $du=dx$. By the change of variable theorem, 

$\int\_{\mathbb{R}}f(x-t)dx = \int\_{\mathbb{R}}f(u)du$. Substituting we have, 

$\int\_{\mathbb{R}}\int\_{\mathbb{R}}f(x-t)g(t)dxdt=\int\_{\mathbb{R}}\int\_{\mathbb{R}}f(u)g(t)dudt$. Since the variables $u$ and $t$ are independent, 

$\int\_{\mathbb{R}}\int\_{\mathbb{R}}f(u)g(t)dudt= \int\_{\mathbb{R}}f(u)du\int\_{\mathbb{R}}g(t)dt$.
Define $\varphi:\mathbb R^2\to\mathbb R^2$ by $(x,t)\mapsto (x-t, t)=:(u,v)$, then the Jacobian is
$$|J| = \begin{vmatrix}\frac{\partial u}{\partial x} & \frac{\partial u}{\partial t}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial t} \end{vmatrix} = \begin{vmatrix}1 & -1\\ 0 & 1\end{vmatrix} = 1. $$
It follows that
$$\int\_{\mathbb R}\int\_{\mathbb R}f(x-t)g(t)\ \mathsf dt\ \mathsf dx = \int\_{\mathbb R}\int\_{\mathbb R}|J|f(u)g(v)\ \mathsf d u\ \mathsf d v = \left(\int\_{\mathbb R} f(x)\ \mathsf dx\right) \left(\int\_{\mathbb R} g(x)\ \mathsf dx\right). $$