Integral of an Odd Function on $\Bbb{R}^{n}$ Given an odd integrable function $\Omega$ on $\Bbb R^n$, i.e. $\Omega \in L^1(\Bbb R^n)$ and $\Omega(-x) = -\Omega(x)$, how do I show that its integral over a symmetric set limited $C$ is zero, ie, if $C = - C$, then
$$\int_{C} \Omega(x)dx = 0.$$
It seems reasonable this assertion, since its true on the real line, but I'm in a little trouble to prove that...
 A: Multiplication by $-1$ gives a regular enough function $g: C \rightarrow C$ to apply the substitution rule:
$$\int_{g(C)} \Omega(x) dx= \int_C \Omega(g(y)) | \det ( g’(y))| dy.$$
Since $g’(y)=g$ because $g$ is linear, and since $g(C)=C$, the above gives
$$\int_{C} \Omega(x) dx =\int_C \Omega(-x) dx = - \int_C \Omega(x) dx.$$
I think you can take it from here.
A: I think the most natural way to do so would be to establish a symmetric partition $P$ on $C$ such that for each $x_i\in P, -x_i\in P$ and then show that the Darboux sums $$U_{\Omega,P}=\sum_{x_i}M_i1_{C}(x^*_i)\Delta x_i=0$$ for all symmetric partitions, $P$. and $$L_{\Omega,P}=\sum_{x_i}m_i1_C(x^*_i)\Delta x_i=0,$$ Then since $$0=U_{\Omega,P}\geq \int_C\Omega(x)dx\geq L_{\Omega,P}=0$$ We can conclude that $\int_C\Omega(x)dx=0$
A: I would like to offer a different basic approach in my opinion. You can approximate your function by symmetric simple functions, for which this statement would be simpler to prove. For $n,m\in \mathbb{N}$, define
$$ E_{m,n}:=\{ x\in C: \frac{m-1}{2^n} \leq \Omega(x)\leq \frac{m}{2^n} \} $$
One can see that
$$ \{ x\in C: -\frac{m-1}{2^n} \geq \Omega(x)\geq -\frac{m}{2^n} \}=-E_{m,n}. $$
Define
$$ \Omega_n(x)= \sum_{m=1}^{\infty} \Big(\frac{m-1}{2^n}\cdot\ \mathbf{1}_{E_{m,n}}(x)-\frac{m-1}{2^n}\cdot\ \mathbf{1}_{-E_{m,n}}(x)\Big) .$$
Then $\Omega_n\to\Omega \cdot \mathbf{1}_C$ almost everywhere and $\vert \Omega_n(x)\vert\leq \vert \Omega(x) \vert$, so
$$ \int \Omega_n(x)dx \to \int \Omega(x)\cdot \mathbf{1}_C(x)dx  $$
by the dominated convergence theorem. Since $E_{m,n}$ and $-E_{m,n}$ have the same measure, we conclude that $\int \Omega_n(x)dx \equiv0$ for all $n$.
