Equal integrals of differential forms implies equal differential forms

Question

How I can prove that if $\alpha$ and $\beta$ are two $r$-dimensional continuous forms defined in an open $U\subset\mathbb{R}^n$ such that $\int_M\alpha=\int_M\beta$ for all compact $r$-dimensional submanifold with boundary $M\subset U$, then $\alpha=\beta$?

Answer 1

Looking at $\omega = \alpha - \beta$ it suffices to prove $\omega$ vanishes on $U$ assuming it integrates to zero on all submanifolds. Let's assume we already have the case $r=n$ figured out. Then the case $r<n$ follows immediately, since for any chart $(V,\varphi)$ of any manifold we have $$\int_{\varphi(V)}(\varphi^{-1})^*\omega = \int_V\omega = 0$$ Hence by the same logic $(\varphi^{-1})^*\omega$ integrates to zero on any open subset of $\varphi(V)\subseteq \mathbb R^r$, so (by what we haven't proved yet) $(\varphi^{-1})^*\omega=0$ on $\varphi(V)$ and thus $\omega = 0$ on $V$ when restricted to input vectors in $T_pV$ (meaning $\omega=0$ as a $r$-form on $V$ and not $U$).

So for any point $p \in U$ and $v_1,\dots, v_r \in T_p\mathbb R^n$ we can take an $r$-dim manifold such that $v_1,\dots, v_r \in T_p M$ (for example, an affine subspace) and using some chart of it around $p$ it shows that $\omega_p(v_1,\dots, v_r)=0$. Since we can do this for all points and input vectors, it follows $\omega = 0$ as an $r$-form on $U$.

So now we can assume $r=n$. But this case is easy, since using local coordinates we have $\omega = f\text{d}x^1 \wedge\cdots\wedge \text{d}x^n$. But $f$ must vanish since if we were to take balls of radius $\epsilon$ around $p\in U$ by the integral mean value theorem $$ f(p) = \lim_{\epsilon\rightarrow 0} \frac{1}{\text{vol}(B(\epsilon,p))}\int_{B(\epsilon,p)}f = \lim_{\epsilon\rightarrow 0} \frac{1}{\text{vol}(B(\epsilon,p))}\int_{B(\epsilon,p)}\omega = \lim_{\epsilon\rightarrow 0}0=0 $$

Edit: I think concluding $\omega = 0$ on $V$ wasn't true. I was confused because this is true when thinking of $\omega|_V \in \Omega^r(V)$ as a form on a manifold and not the ambient open neighborhood. But infact, by definition $(\varphi^{-1})^*\omega$ being zero translates only to how $\omega$ acts on tangent vectors to $V$ at $p$. This is not a major problem and I have updated my solution.