prove that $f_n\in L^1_\text{loc}(\mathbb{R})$

Question

Today I have a exercise about locally integral as following : Defined ($n\in\mathbb{N^{*}}$ :

$$f_{n}(x)=\begin{cases}\log |x| , &|x|>\frac{1}{n} \\ -\log n , &|x|<\frac{1}{n}\end{cases}$$

Question :prove that $f_n\in L^1_\text{loc}(\mathbb{R})$

I think we can solve it directly by definition :

$$f_n\in L^1_\text{loc}(\mathbb{R})\iff \int_{\mathbb{R}}|f_n(x)| \, dx<\infty $$

Now let $[a,b]\in\mathbb{R}$ mean study three cases first Cases

$1- a>0$ , $h\to 0$ $$\int_h^{\frac{1}{n}} \left|-\log n\right|\,dx +\int_{\frac{1}{n}}^a|\log |x|| \,dx <\infty $$

$2- b<0$ same way

Now $0\in [a,b]$ I don't know how My another idea use convergent domain mean $$f_n \overset{{P.P}} \to f$$

Then :

$$|f_n|≤g \in L_\text{loc}^1 $$

But how i complete first and second method?

Answer 1

For $|x|>\frac{1}{n}$, $f_n$ is continuous so it is locally bounded there and thus locally integrable. For $|x|<\frac{1}{n}$ is of course bounded and thus locally integrable. Now you can always break down every interval or measurable set in disjoint sets such that $|x|>\frac{1}{n}$ on the one subset and $|x|<\frac{1}{n}$ on the other.

Answer 2

It seems like the fastest way is noting that $f_n$ (extended in the obvious way to $\pm 1/n$) is continuous. Every continuous function is locally integrable.