Lebesgue Theory in Higher dimensions

What are some things that we have to watch out for when going to higher dimensions (greater than or equal to 2) for Lebesgue Measure Theory?

For instance, is there anything that is true in Lebesgue Measure Theory for $\mathbb{R}$ but not $\mathbb{R}^n$ or vice versa?

• If you stay with Borel sets, beside that $1/|x|$ becomes integrable, it shouldn't be very different. Are you interested about the worst cases, for example the measure of fractal sets ? – reuns Jul 12 '16 at 9:37
• @user1952009 Interesting. Why would $1/|x|$ be integrable? – yoyostein Jul 12 '16 at 9:44
• I meant locally integrable. because $f(x) = \left\lfloor \frac{1}{|x|} \,\right\rfloor$ is : $\displaystyle\int_{|x| < 1} \left\lfloor \frac{1}{|x|} \,\right\rfloor dx = \sum_{n=1}^\infty \mu(A_n)$ where $A_n = \{ x \ \mid \ \left\lfloor \frac{1}{|x|} \,\right\rfloor =n \ \}$, i.e. $A_n = \{ x \ \mid \ \frac{1}{n+1} < |x| \le \frac{1}{n} \ \}$ and $\mu(A_n) = ?$ – reuns Jul 12 '16 at 9:51
• I meant $\int_{|x| < 1} f d\mu$ but you understood it. In $\mathbb{R}^k$, an annulus $1/(n+1) < |x| < 1/n$ should have a volume $R_k (n^{-k}-(n+1)^{-k})$ where $R_k$ is the volume of the unit sphere, so $\mu(A_n) = R_k (n^{-k}-(n+1)^{-k})$. Anyway it is still true in any dimension that a continuous function $f$ is Lebesgue integrable iff $|f|$ is improper Riemann integrable, and with a radial change of variable : $\displaystyle\int_{|x| < 1} \frac{dx}{|x|^a} = R_k \int_0^1 \frac{r^{k-1}}{r^a} dr$ – reuns Jul 12 '16 at 10:00
• The Banach-Tarski paradox only works in $\mathbb R^n$ for $n\geq 3$. In $\mathbb R$ and $\mathbb R^2$ it's a fact that if you can write $A$ as the disjoint union of $A_1,\dots,A_k$ and $B$ as the disjoint union of $B_1,\dots,B_k$ where each $A_i$ is a rigid transformation of $B_i$, then if both $A$ and $B$ are measurable they have the same measure. – Oscar Cunningham Jul 13 '16 at 9:18