On a differentiable manifold $M$ there is a standard notion of zero measure sets using charts ( a set $A\subseteq M$ has zero measure if for every chart $(U,\varphi)$ $\varphi(A\cap U)$ has zero Lebesgue measure in $\mathbb{R}^n$)
Moreover if we have a volume form $\Omega$ on $M$, by the Riesz-Kakutani representation there is a unique Radon measure $\mu$ on the Borel $\sigma$-algebra on $M$ such that for all compactly supported continuous function on $M$ we have \begin{equation*} \int_M f \Omega=\int_M f \mu. \end{equation*} How can I prove that if A is a Borel set with zero measure in the chart definition then it is also $\mu(A)=0$?
Is this fact actually true? I think so but I haven't found any prof yet. I tried for example using regularity of $\mu$ and approximation of characteristic functions with bump functions...
Thank you in advance.
Here's the argument adapted to the Riesz definition of the measure. For each point $p\in A$, choose a positively oriented chart $(U_p,\alpha_p)$ such that $p\in U_p$ and $\Omega|_{U_p}=\Omega_{\alpha_p}\,dx_{\alpha_p}^1\wedge \cdots\wedge dx^n_{\alpha_p}$ for some smooth bounded positive function $\Omega_{\alpha_p}:U_p\to (0,\infty)$ (positivity is because of the positively-oriented chart, and boundedness can be assured by shrinking the set $U_p$ small enough). We can now extract a countably many such $U_p$ whose union covers $A$. Then, $\mu(A)\leq \sum\mu(U_p\cap A)$, the sum over countably many $U_p$'s. If we show each of the terms on the right vanishes, then $\mu(A)=0$.
Hence, we are now reduced to the following situation: there is a single chart $(U,\alpha)$ such that $A\subset U$, $\lambda_n(\alpha[A])=0$, $\Omega=\Omega_{\alpha}\,dx^1\wedge \cdots\wedge dx^n$ on $U$, and $\Omega_{\alpha}:U\to (0,\infty)$ is smooth, positive and bounded, say by a constant $B>0$. Now, by the regularity properties of $\mu$ (or really if you just look at the proof of Riesz's theorem), we have \begin{align} \mu(A)&=\inf_{\substack{\text{$V\subset U$ open}\\A\subset V}}\mu(V)= \inf_{\substack{\text{$V\subset U$ open}\\A\subset V}}\sup_{f\prec V}\int_M f\Omega. \end{align} So we take infimum over all open sets $V\subset U$ which contain $A$, and the supremum of the integrals over all continuous $f$ with $0\leq f\leq 1$ and $\text{supp}(f)\subset V$. Now, why did I go through all the trouble with the charts? it is so that I can now evaluate the integral on the right in a very simple manner. Recall that integrals of compactly supported continuous $n$-forms is defined by a sum involving a partition of unity; but if we have the form supported inside of a single positively-oriented coordinate chart, then the integral simplifies the most obvious thing, i.e integrate the chart-representative function \begin{align} \int_Mf\Omega&=\int_{\alpha[U]}(f\cdot\Omega_{\alpha})\circ\alpha^{-1}\,d\lambda_n. \end{align} We now use the obvious estimate $f\leq \chi_V$ and $\Omega_{\alpha}\leq B$ to get \begin{align} \int_Mf\Omega\leq \int_{\alpha[U]}(\chi_V\circ \alpha^{-1})\cdot B\,d\lambda_n=B\int_{\alpha[V]}1\,d\lambda_n=B\cdot\lambda_n(\alpha[V]). \end{align} Therefore, \begin{align} \mu(A)&\leq \inf_{\substack{\text{$V\subset U$ open}\\A\subset V}}B\lambda_n(\alpha[V])=B\cdot\inf_{\text{$V'\subset \alpha[U]$ open}\\ \alpha[A]\subset V'}\lambda_n[V']=B\cdot \lambda_n(\alpha[A])=0. \end{align} The first equality is trivial, I'm just using the homeomorphism $\alpha$ to 'transfer' sets on the manifold to $\alpha[U]$. The second equality is because the $n$-dimensional Lebesgue measure is outer-regular, and the final equality is by hypothesis on $A$. Hence, $\mu(A)=0$. This completes the proof.