# Why This Exachange of Integral Works?

I’m reading Evan’s PDE. And I got stuck in the proof of properties of mollifier(Pg. 715). The property is that:

(iv) If $$1\leq p<\infty$$ and $$f\in L^p_\text{loc}(U)$$, then $$f^\varepsilon \to f$$ in $$L^p_\text{loc}(U)$$.

at the step 4 attempting to control $$\lVert f^\varepsilon\rVert_{L^p(V)}$$ by $$\lVert f\rVert_{L^p(W)}$$($$V\subset\subset W\subset\subset U$$):

$$$$\begin{split} \int_V|f^\varepsilon(x)|^p \, dx&\leq\int_V\left(\int_{B(x,\varepsilon)} \eta_\varepsilon(x-y) |f(y)|^p\ dy\right)\ dx\\ &\leq\int_W|f(y)|^p \left(\int_{B(y,\varepsilon)}\eta_\varepsilon(x-y)\ dx\right)\ dy=\int_W|f(y)|^p\ dy, \end{split}$$$$ provided $$\varepsilon>0$$ is sufficiently small.

It is clear that condition “$$\varepsilon\to 0$$” is used to make the second ‘ $$\leq$$’ happen. But I don’t know how. Is Fubini’s theorem or LDC applied?

Edit: Maybe I should post the entire proof to see more clear: proof of Step 4

Since $$V\subset \cup_{x\in V} B(x,\epsilon) \subset W$$ if $$\epsilon$$ small enough, we can prove: $$\{(x,y): x\in V , y\in B(x,\epsilon)\} \subset \{(x,y): y \in W, x\in B(y,\epsilon)\}.$$

Proof:

$$\forall (x,y) \in \{(x,y): x\in V , y\in B(x,\epsilon)\}$$, 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 "$$\subset$$".

Before applying Funibi, we conclude $$\int_{ \{(x,y): x\in V , y\in B(x,\epsilon)\}} \le \int_{\{(x,y): y \in W, x\in B(y,\epsilon)\}}$$ given the fact that the integrand is nongegative.

• Thanks! Your proof is very concise, and I intend to post some graphics to illustrate the containing relationship:) – Niacrosis Hermit Oct 25 at 6:38

For positive integrands, many call this Tonelli's Theorem, which doesnt even require integrability to change order of integration. The reason for $$\varepsilon$$ small is so that $$B(y, \varepsilon)$$ is inside $$W$$ for all $$y \in V$$.

• Oh I see it. And how can $B(x,\epsilon)$ be transformed to $B(y,\epsilon)$? – Niacrosis Hermit Oct 25 at 3:47

The diagram is a complement of yuguaw’s answer. The green area is $$V_\epsilon=\{(x,y)|x\in V, y\in B(x,\epsilon)\}$$ and the blue one is $$W_\epsilon=\{(u,y)|y\in W, u\in B(y,\epsilon)\}$$.

schematic diagram