Show that $f:\Bbb R^n\to\Bbb R$, given by: $$ f(x) = \begin{cases}\sin\left(\frac{1}{\|x\|}\right)\|x\|^{-n-\arctan(\|x\|-1)} & x\not=0 \\ 0 & x=0 \\ \end{cases}$$ is Lebesgue integrable.

I don't know how to prove this, I want to use a corolary of the dominated convergence theorem, but I just got really confused. I considered the family of rings given by $A_k:=\{x\in\Bbb R^n:\frac{1}{k}\le \|x\|\le k\}$, then $A_k\subset A_{k+1}$ and $\cup_{k\in\Bbb N} A_k=\Bbb R^n-\{0\}$. Now if we consider the function $$f|_{A_k}(x)=\begin{cases} \sin\left(\frac{1}{\|x\|}\right)\|x\|^{-n-\arctan(\|x\|-1)} & \|x\|\in[1/k,k] \\ 0 & \|x\|\not\in[1/k,k] \end{cases}$$ then is well defined and we want it to be Lebesgue-integrable in $A_k$ for any $k\in\Bbb N$, however I don't know how to do this.

There's also this other exercise, that says:

Let $X_1,X_2$ be integrable in $\Bbb R^n$ such that $X_1\cap X_2$ is null. If $X=X_1\cup X_2$ then $f:X\to\Bbb R\cup\{\pm\infty\}$ is integrable in $X$ $\iff$ $f|_{X_i}$ is integrable in $X_i$ for $i=1,2$, and $\int_X f=\int_{X_1}f+\int_{X_2}f$

I think that is almost the same problem, we have to show that the restricted function is integrable. I'm really lost with this problem...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.