Inequality Involving $\int_V \left(\int_{B(x,\epsilon)} \eta_{\epsilon}(x-y) |f(y)|^p dy\right) dx $ I was reading the book "Partial Differential Equation" written by Lawrence C. Evans, coming up with a question.
On page 718, Evans wrote

$$\int_V \left(\int_{B(x,\epsilon)} \eta_{\epsilon}(x-y) |f(y)|^p dy\right) dx \\ \leq \int_W |f(y)|^p \left(\int_{B(y,\epsilon)} \eta_{\epsilon}(x-y) dx\right) dy$$

Where $V,W,U$ are  open sets, $V\subset\subset W\subset \subset U$, and $\eta_{\epsilon}(x) := \frac{1}{\epsilon^n} \eta(\frac{x}{\epsilon})$, $\eta$ is the standard mollifier, $f \in L_{loc}^p(U)$
I want to ask why the inequality holds. I was trying to prove it so hard but I have no ideas.
 A: This is an application of Fubini's theorem, followed by the fact that the integrand is positive and $V$ is a subset of $W$.
A: Choise some small $\varepsilon>0$ such that $B(x,\varepsilon)\subset V$, we have
$$\int_V\bigg(\int_{B(x,\varepsilon)}\eta_\varepsilon(x-y)|f(y)|^pdy\bigg)dx=\int_V\bigg(\int_V\eta_\varepsilon(x-y)|f(y)|^p~\chi_{B(x,\varepsilon)}(y)~dy\bigg)dx$$
where $\chi_{B(x,\varepsilon)}(y)$ is characteristic function.
Now, using the Fubini's theorem,
$$\int_V\bigg(\int_V\eta_\varepsilon(x-y)|f(y)|^p~\chi_{B(x,\varepsilon)}(y)~dy\bigg)dx=\int_V|f(y)|^p\bigg(\int_V\eta_\varepsilon(x-y)~\chi_{B(x,\varepsilon)}(y)~dx\bigg)dy$$
fixed variable $y$, the characteristic function $\chi_{B(x,\varepsilon)}(y)=1$ only at $x\in B(y,\varepsilon)$, thus
$$\int_V|f(y)|^p\bigg(\int_V\eta_\varepsilon(x-y)~\chi_{B(x,\varepsilon)}(y)~dx\bigg)dy=\int_V|f(y)|^p\bigg(\int_{B(y,\varepsilon)}\eta_\varepsilon(x-y)dx\bigg)dy.$$
Finally, since the fact that the integrand is positive, then
$$\int_V|f(y)|^p\bigg(\int_{B(y,\varepsilon)}\eta_\varepsilon(x-y)dx\bigg)dy\le \int_W|f(y)|^p\bigg(\int_{B(y,\varepsilon)}\eta_\varepsilon(x-y)dx\bigg)dy.$$
A: Since $V\subset \cup_{x\in V} B(x,\epsilon) \subset W$ if $\epsilon$ small enough,  we can show:
$\{(x,y): x\in V , y\in B(x,\epsilon)\} \subset \{(x,y):   y \in W, x\in B(y,\epsilon)\}.$
$\forall (x,y) \in \{(x,y): x\in V , y\in B(x,\epsilon)\}$, we know $x \in V$, then $y \in B(x,\epsilon)\subset \cup_{x\in V} B(x,\epsilon)\subset W$. And $y \in B(x,\epsilon)$ imples $x\in B(y,\epsilon)$. Therefore $(x,y) \in \{(x,y):   y \in W, x\in B(y,\epsilon)\}$, which proves "$\subset$".
Then by the above fact and the Fubini Thm,
$\int_V \int_{B(x,\epsilon)} (\cdot)dy\,dx = \int_{ \{(x,y): x\in V , y\in B(x,\epsilon)\}}(\cdot)dy\, dx \le  \int_{\{(x,y):   y \in W, x\in B(y,\epsilon)\}}(\cdot) dy\,dx = \int_{\{(x,y):   y \in W, x\in B(y,\epsilon)\}}(\cdot) dx\,dy = \int_W \int_{B(y,\epsilon)}(\cdot)dx\,dy$ 
