Estimating the integral of convolution by a product of integrals I'm reading the appendix of Evans' PDE book. One part of it says if $V$ is a compact subset of $\mathbb R^n$ and $\phi_\epsilon$ from $\mathbb R^n$ to $[0,\infty)$ has support inside $B(0,\epsilon)$. $f$ is an integrable function on $\mathbb R^n$. And $W$ contains all points in $\mathbb R^n$ with distance less or equal than $\epsilon $ to $V$.Then $$\int _V \int_ { B(x, \epsilon )} \phi_\epsilon (x - y) |f(y)| dy dx \leq \int _W |f(y) | \int _{B(y,\epsilon ) } \phi_\epsilon (x -y ) dx dy$$
I want to know how can I understand this inequality? I tried to see it through Fubini's theorem but I failed as the domains in the Fubini's theorem are not defined through functions of $x$ or $y$. 
 A: You can remove the dependence of the domains of integration from $x$ resp. $y$ by multiplying the integrand with the characteristic function of the domain, and extending the integrals over $\mathbb{R}^n\times \mathbb{R}^n$. However, I think it is more instructive to let that be and directly look at the domains of integration. For the left hand side, we integrate over
$$A = \{ (x,y) \in \mathbb{R}^n\times \mathbb{R}^n : x\in V, \lVert x-y\rVert < \epsilon\}.$$
For the right hand side, we integrate over
$$B = \{ (x,y) \in \mathbb{R}^n \times \mathbb{R}^n : y \in W, \lVert y-x\rVert < \epsilon \}.$$
Since the integrand $g(x,y) = \phi_\epsilon(x-y)\cdot \lvert f(y)\rvert$ is nonnegative, the inequality directly follows from the inclusion $A \subset B$. But that inclusion is easy to see from the definition of $W$: If $(x,y) \in A$, then $x\in V$ and $\lVert x-y\rVert < \epsilon$, whence $\operatorname{dist}(y,V) \leqslant \lVert x-y\rVert < \epsilon$, so $y\in W$, and thus $(x,y) \in B$.
(Make the inequalities non-strict if $B(x,\epsilon)$ denotes the closed ball of radius $\epsilon$ and not the open ball.)
