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Suppose $Q$ is a rectangle in $\mathbb R^n$ and $f: Q \to \mathbb R$ is a bounded function. How is it that if $f$ vanishes outside a closed set $B$ of measure zero, the integral of $f$ sub $Q$ exists and equals zero?

I can introduce $D$ to be the set of points of $Q$ at which $f$ fails to be continuous and then integral of $f$ sub $Q$ exists and equals zero. But how to make this rigorous?

my proof:

Select a refinement, B, of Q s.t., we are interested in rectangles that don't intersect B, and which are those that have measure zero.

The squares that don't indersect at all, have no discontinuities. We note that discontinuity occurs if we are to select B intersect Q sub j is a subset of B. B has measure zero, so B intersect Q sub j has measure zero and j is an arbitrary integer.

B is closed implies B intersect Q closed and contains all discontinuity points.

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Are the integrals you speak about Lebesgue integrals? If so, then squeezing $|f|$ between the zero function and some multiple of the indicator function of $B$ should work. – Henning Makholm Feb 16 '12 at 11:06
$f$ doesn't even have to be bounded – user8268 Feb 16 '12 at 11:56
I think so, but need help in writing it out – James R. Feb 16 '12 at 21:23

Well you are speaking about the Riemann integral. For solve that you should find a partition of rectangles such that $B\subset \cup_{i} B_{i}$ where $\sum_{i}|B_{i}|<\frac{\epsilon}{\sup f+1}$ then the superior sum will be lesser than $\epsilon$ for any $\epsilon>0$, what means that $\bar{\int f}\leq0$.

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