Finite Integral in measure theory

[scottneh]

Hello, I am preparing for a screening exam and I'm trying to figure out some old problems that I have been given.

Given:

Suppose $$\mu$$ is a finite Borel measure on R, and define

f$$(x)=\int\frac{d\mu(y)}{\sqrt{\left|x-y\right|}}$$

Prove f$$(x)$$ is finite almost everywhere

If I integrate I get -2$$\sqrt{x-y}$$ for y$$\leq$$x and 2$$\sqrt{-x+y}$$ for x$$\le$$y

I can clearly see that the integral is not finite for x = y and hence the "almost everywhere constraint and the integrand looks like it would tend to zero as either variable increases toward infinity but the result of the integral looks like it will blow up if wither variable tend to infinity.

I was looking at Fubini-Tonelli but I'm not really sure how to use it here as only d mu(y) and not a d mu(x) as well.

Can someone please help me get started?

Thanks

 

[jvicens]

for reaching out for help with this problem! it's important to have a strong understanding of mathematical concepts and techniques, so it's great that you are taking the time to work through these old problems.

To prove that f(x) is finite almost everywhere, we need to show that for almost all x, the integral \int\frac{d\mu(y)}{\sqrt{\left|x-y\right|}} is finite.

First, let's consider the case when x is a fixed point. In this case, the integral becomes \int\frac{d\mu(y)}{\sqrt{\left|x-y\right|}} = \int\frac{d\mu(y)}{\sqrt{\left|x-x\right|}} = \int\frac{d\mu(y)}{0}, which is undefined. However, this case is irrelevant for the almost everywhere constraint, as it only concerns a single point.

Next, let's consider the integral for a fixed y, where y is a point in the support of \mu. In this case, the integral becomes \int\frac{d\mu(y)}{\sqrt{\left|x-y\right|}} = \int\frac{d\mu(y)}{\sqrt{\left|y-y\right|}} = \int\frac{d\mu(y)}{0}, which is again undefined. However, this case is also irrelevant for the almost everywhere constraint, as it only concerns a single point in the support of \mu.

Now, let's consider the integral for a fixed x outside of the support of \mu. In this case, the integral becomes \int\frac{d\mu(y)}{\sqrt{\left|x-y\right|}} = \int\frac{d\mu(y)}{\sqrt{\left|x-y\right|}}, which is finite since x is outside of the support of \mu and the integrand is well-defined.

Finally, let's consider the integral for a fixed x inside the support of \mu. In this case, the integral becomes \int\frac{d\mu(y)}{\sqrt{\left|x-y\right|}} = \int\frac{d\mu(y)}{\sqrt{\left|y-x\right|}}. Now, we can use Fubini-Tonelli's theorem to switch the order of integration, since both x and y are inside the support of \mu. This gives