Studying the integrability of a function $f:\mathbb R^3\to\mathbb R$ in a neighbourhood of its singularity

Question

Today we solved for the first time exercises concerning the Lebesgue integrabilty of functions in $2$ or $3$ variables.
We followed a standard method for every exercise. For example, consider a function $f:\mathbb R^3\to\mathbb R$ with a singularity in the origin and suppose we want to study the integrability on a region $E\subset\mathbb R^3$ s.t. $(0,0,0)\in\mathbb R^3$.
We know that $f\in\mathcal C^0(\mathbb R^3\setminus\{(0,0,0)\})$ so, in order to discuss the convergence of $\iiint _E f(x,y,z)dxdydz$, we considered a set like $\mathbb B_{\varepsilon}(\textbf 0)$, $[-\varepsilon,\varepsilon]^3$ or $[0,\varepsilon]^3$ and we studied (for example) $$\lim_{\varepsilon\to 0}\iiint_{E\setminus \mathbb B_{\varepsilon}(\textbf 0)}f(x,y,z)dxdydz.$$ So I wondered why we did this, I wanted to undertand the reason behind this process. If the functionn is integrable, then the integral of $f$ does not depend to which sets we remove but I don't think we saw a similar result in theory lessons.

Answer 1

To simplify I will take $E = \overline{\mathcal{B}}(0,r)$, you know that $f$ is continuous on $E \setminus \{0\}$ so for each $0 < \epsilon < r$ the integral $$ \int_{E \setminus \mathcal{B}(0,\epsilon)} f d \lambda $$ is the integral of a continuous function on a compact set. It is therefore well defined as a real number. Now you are interested in weather or not the integral $$ \int_E |f| d\lambda$$ is finite. The link between these two integrals is that $$|f| 1_{E \setminus \mathcal{B}(0,\epsilon)} \rightarrow |f|$$ when $\epsilon$ goes to 0, for almost every $x \in E$. This convergence is increasing so by the Beppo-Levi's lemma you have that $$ \int_E |f| d\lambda = \lim_{\epsilon \rightarrow 0} \int_{E \setminus \mathcal{B}(0,\epsilon)} |f| d \lambda.$$ Now you have to bound from above the right hand term with something finite that doesn't depend on $\epsilon$.

If you assume $f$ to be integrable on all $\mathbb{R}³$ then you have automatically $f$ integrable on $E$ and there is no need to introduce those integrals. But in general a function that is merely continuous is not integrable.

Does this answer your question ?

Answer 2

Because of the following theorem:

Let $ A_1 \subseteq A_2 \subseteq \cdots $ be an increasing sequence of measurable sets with union $ A = \bigcup_{n=1}^\infty A_n $. If $$ \lim_{n \to \infty} \int_{A_n} |f(x)| dx < \infty $$ then $f$ is integrable on $A$ and we have $$ \int_A f(x) dx = \lim_{n \to \infty} \int_{A_n} f(x) dx $$

This can be proved by the dominated convergence theorem and I leave it as an exercise.

Now, let $ A_n = E\setminus \mathbb B_\frac{1}{n}(0) $ for each $n \in \mathbb{N}$. Then $ \{A_n\} $ is an increasing sequence of measurable sets with union $ E \setminus \{0\} $. Moreover, we have

$$ \lim_{\epsilon \to 0} \int_{E\setminus \mathbb B_{\epsilon}(0)} |f(x)| dx = \lim_{n \to \infty} \int_{A_n} |f(x)| dx $$

If the above quantity is finite, then by the theorem stated in the beginning, $f$ is integrable on $E$ and satisfies

$$ \lim_{\epsilon \to 0} \int_{E\setminus \mathbb B_{\epsilon}(0)} f(x) dx = \lim_{n \to \infty} \int_{A_n} f(x) dx = \int_{E} f(x) dx $$

because $E$ has the same measure as $ E \setminus \{0\} $.