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Let $f:X\rightarrow R$ for $X \subset R^n$.

I want to prove that \begin{equation} \int_{I} f d\lambda = 0 \end{equation} for all intervals $I \subset R^n$ implies that $f=0$ $\lambda$-almost everywhere on $X$. I guess the proof would be similar to the one here.

Take the rectangle $A_r(x) = \prod_{i=1}^{n}[x_i-r/2,x_i+r/2]$ centered at $x$ with edge length $r$, so $|A_r(x)|=r^n$. By assumption \begin{equation} \frac{1}{r^n} \int_{A_r(x)} f d\lambda = 0 \end{equation} for all $r > 0$.

By Lebesgue differentiation theorem, \begin{equation} f(x) = \lim_{r \rightarrow 0} \frac{1}{r^n} \int_{A_r(x)} f d\lambda = 0 \end{equation} almost everywhere on $X$.

Is this reasoning correct?

Thanks in advance.

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  • See also http://math.stackexchange.com/questions/1030050/let-f-be-measurable-and-a-b-in-mathbbr-with-frac1-lambdam-int-mf?rq=1 – Siminore Sep 07 '16 at 09:38
  • How much is the measure of $I$? – Haipeng Chen Sep 14 '16 at 03:08
  • The measure is the Lebesgue measure, i.e. if $I = I_1 \times \cdots \times I_n$ is the interval in $R^n$, then $\lambda(I) = \prod_{i=1}^{n} \lambda(I_i)$ is its measure. – user353893 Sep 14 '16 at 09:01

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