# Fubini Theorem for measure zero

I know Fubini Theorem in calculus, but the measure zero version does not make sense to me:

$n=k+1$, and $V_c$ is the "vertical slice" {c}$\times R_l$. Let $A$ be a closed subset of $R^n$ such that $A \cap V_c$ has measure zero in $V_c$ for all $c \in R^k$. Then $A$ has measure zero in $R^n$.

Could someone help me explain what does this mean, and how it relates to the Fubini Theorem in calculus?

Thank you very much!

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Can you explain what the $V_c$ are? –  Rookatu May 30 '13 at 5:55
Hi @Rookatu thank you very much for your reminder. $V_c$ is the "vertical slice" {c}$\times R_l$. Also, $n=k+1$. –  1LiterTears May 30 '13 at 6:55

how it relates to the Fubini Theorem?

Let $\chi_A$ be the characteristic function of $A$: that is, $\chi_A(x)=1$ when $x\in A$ and $0$ otherwise. The integral of $\chi_A$ with respect to whatever measure is equal to the measure of $A$. This is why the characteristic function is used to make the transition from measures to integrals and back.

Fubini's theorem is about integrals. Applied to $\chi_A$, it says that the integral of $\chi_A$ can be found by integrating over each vertical slice first, and then integrating over $c$. Well, if every slice has measure zero, then the integral over every slice is zero. Then $\int 0\,dc =0$, and the conclusion is that $A$ has measure zero.

what does this mean?

Think of the calculus exercise about finding volume by integrating the area of cross-sections. If the area of each cross-section happened to be zero, the volume is zero. That's all there is to it, but a rigorous proof takes some effort, mostly to make sure that the concepts are properly defined.

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That makes so much sense and is so clear! Thank you very much! –  1LiterTears May 30 '13 at 17:59