Lebesgue Integral, an example

Question

I have been trying to studying the construction of Lebesgue integral for a while now. I am following the Princeton Lectures on Analysis and I am stuck at the part where it defines the integral of non-negative functions. I find the definition to be quite clear but I cannot understand the examples given here.

With the above definition of the integral, there are only two possible cases; the supremum is either finite, or infinite. In the first case, when $\int f(x)dx<+\infty$, we shall say that $f$ is Lebesgue integrable or simply integrable.

Clearly, if $E$ is any measurable subset of $\mathbb R^d$, and $f\ge0$, then $f_{\chi_E}$ is also positive, and we define $$\int f(x)dx=\int f(x)\chi_E(x)dx.$$ Simple examples of functions on $\mathbb R^d$ that are integrable (or non-integrable) are given by $$f_a(x)=\begin{cases}|x|^{-a}&\text{ if }|x|\le1,\\0&\text{ if }|x|>1.\end{cases}$$ $$F_a(x)=\frac1{1+|x|^a}, \text{ all }x\in\mathbb R^d.$$ Then $f_a$ is integrable exactly when $a<d$, while $F_a$ is integrable exactly when $a>d$.

How are the values of "a" here in these two examples making the function integrable or not !? and what does the value of "a" has to do with "d" ?

Answer 1

Consider a map $f : \mathbb R^d \to \mathbb R$ which is only dependent on the radius $R$ in the $n$-sphere Spherical coordinates. I.e. $f(x)=g(\vert x\vert)$ where $g$ is a real map.

You then have

$$\int_{\mathbb R^d} f(x) \ dx = K_d \int_0^\infty g(R) R^{d-1}\ dR$$ where $K_d$ is a constant that only depends on $d$. This uses an integral by substitution in dimension $d$.

You can use that to look at the integrability of the maps of your original question.

Answer 2

Likely this is explained further down in your text. Anyway...

The Lebesgue integral of $f_a$ over $\mathbb R^d$ is: $$\int_{\mathbb R^d} f_a\,d\mu = \int_0^\infty \mu\big(\{x\in B_d(1):f_a(x)>y\}\big)\,dy = \int_0^\infty \mu\big(\{x\in B_d(1):|x|^{-a}>y\}\big)\,dy$$ where $B_d(r)$ is the unit ball in $d$ dimensions with radius $r$.

If $a\le 0$ there is no singularity nor infinity involved, and the result is finite.

If $a>0$ then the measure at $y$ is the volume of the $d$-ball $B_d(r)$ with radius $r=y^{-\frac 1a}\le 1$. That also means that $y\ge 1$. The volume of $B_d(r)$ is $C_d r^d$ for some constant $C_d$ that depends only on $d$ (for instance $C_2=\pi$ and $C_3=\frac 43\pi$). So the integral becomes: $$\int_{\mathbb R^d} f_a\,d\mu = \int_1^\infty C_d\cdot(y^{-\frac 1a})^d\,dy = \frac{C_d}{-\frac da+1} y^{-\frac da+1}\Bigg|_1^\infty $$ This is finite iff $-\frac da+1<0 \iff a<d\quad$ (for the case $a>0$).

Therefore $f_a$ is Lebesgue integrable over $\mathbb R^d$ iff $a<d$.