If $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is a locally integrable function and $\int_{[a,b]^n}f=0$ for all reals $a<b$, is it true that $f=0$ almost everywhere?
By Lebesgue Differentiation Theorem I know that $f(t,\stackrel{n)}{\ldots},t)=0$ for all $t\in\mathbb{R}$, but I am not sure about what happens for a general point in $\mathbb{R}^n$.
Notation: $[a,b]^n=[a,b]\times \stackrel{n)}{\cdots}\times [a,b]$; and $\int$ is a Lebesgue integral.
You can't conclude this in $n \ge2$ dimensions, consider e.g. $f(x_1, \ldots, x_n) = x_1 - x_2$.
More generally, you can choose $f(x) = g(x_1) - g(x_2)$ for some sufficiently nice function $g$ to get functions $f$ that satisfy stronger conditions than just local integrability (i.e. smoothness, compact support, etc.).