# 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.

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