Htykuut

Let $(mathbb R, mathcal{B}, lambda) $ be a measurable space, where $mathcal{B}$ denotes the Borel sigma algebra on $mathbb R$ and $lambda$ the Lebesgue measure.

Is the function :

$$f(x)= begin {cases} +infty &text { if } x=0\ ln|x| &text { if }0<|x|<1\0 &text{ if }|x|ge 1 end {cases}$$

Lebesgue integrable ?

It is my first exercise of that type and I don't know how to proceed.

This question is relatively simple, but doesn't deserve so many downvotes. First note that $ln(x)$ is negative on $(0,1)$. Thus we need to show the integrability of $|ln(x)|$. Because of the continuity, the Lebesgue-integral over $(varepsilon,1)$ is the same as the Riemann-integral. So, we only need to show that this function is Riemann-integrable. (Apply monotone convergence theorem with $varepsilon downarrow 0$.)

Now note that $f(x) = x ln(x)$, $x in (0,infty)$, can be continuously extenend on $[0,infty)$ with $f(0)=0$. Thus
$$int_varepsilon^1 ln(x) , d x = x ln(x) Big|_{x= varepsilon}^1 - int_varepsilon^1 1 , d x = -varepsilon ln(varepsilon)-(1-varepsilon).$$
For $varepsilon downarrow 0$ we get that this term convergences towards $-1$.