# Ngô Quốc Anh

## August 31, 2008

### L^1 hội tụ hầu khắp nơi

Suppose . Show that the integral

$\phi(x)=\int_{\mathbb{R}^3}\frac{f(y)dy}{|x-y|}$

converges for a.e. .

Solution. Let

$\psi(x)=\int_{\mathbb{R}^3}\frac{|f(y)|}{|x-y|}\,dy.$

Let  be the ball centered at the origin of radius 

$\int_{B_r}\psi(x)\,dx\le\int_{B_r}\int_{\mathbb{R}^3} \frac{|f(y)|}{|x-y|} \,dy\,dx$ $=\int_{\mathbb{R}^3}|f(y)|\int_{B_r}\frac{1}{|x-y|}\,dx\,dy$

We can show that

$\int_{B_r}\frac{1}{|x-y|}\,dx\le\int_{B_r}\frac{1}{|x|}\,dx= Cr^2.$

(The constant  can be computed – it’s actually  – but the exact value is immaterial.) Hence,

$\int_{B_r}\psi(x)\,dx\le Cr^2\int_{\mathbb{R}^3}|f(y)|\,dy$.

Since this is finite, we must have  almost everywhere on  Since we can repeat this for any  (choose a countable sequence of such  tending to ) we can say that  is finite almost everywhere on  Hence the integral that defines  converges absolutely for almost every  Note the importance here of having the function  be locally integrable – in fact, uniformly locally integrable.