Show that $f=0$ almost everywhere. If $f$ is integrable in $\mathbb{R}^d$ as for the Lebesgue measure and $\int_{R}f=0$ for each rectangle $R$, then $f=0$ almost everywhere.
Could you give me some hints how to show it??
 A: You can try the following step:


*

*show that for each positive $\varepsilon$, there exists a $\delta$ such that if $B\subset\mathbb R^d$ is a Borel subset of Lebesgue measure smaller than $\delta$, then $\int_B|f|\mathrm d\lambda_d\lt\varepsilon$;

*if $B$ is a Borel subset of finite measure, there exists a finite union of rectangles $B'$ such that $\lambda_d(B\Delta B')\lt\delta$. Assume first it is contained in $[-R,R]^d$ for some $R$, and use the fact that in a finite measure space, each measurable set can be approximated for the metric $d(A,B):=\mu(A\Delta B)$ by an element of a generating algebra. Here, you will need to use the algebra of finite unions of rectangles. 

A: This follows from the Lebesgue differentiation theorem, if it is available.
For every $x\in\mathbb R^n$, let $(R_k(x))_{k\ge0}$ be a sequence of rectangles with positive measure tending to $\{x\}$ (e.g. $R_k(x)=[x_1-1/k,x_1+1/k]\times\dots\times[x_n-1/k,x_n+1/k]$).
Since the family of rectangles of $\mathbb R^n$ is of bounded eccentricity (in the sense of the Wikipedia page), for a.e. $x\in\mathbb R^n$,
$$
f(x)=\lim_{k\to\infty}\frac1{\left|R_k(x)\right|}\int_{R_k(x)}f(t)\,\mu(\mathrm dt)=0.
$$
