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## Is the 2d gauge integral equivalent to the Lebesgue integral for nonnegative functions?

Let $f$ be a function from $[0,1]\times [0,1]$ to $\mathbb{R}$.

Definition:

2dgauge$\displaystyle\int f \; = \; I$
$\Leftrightarrow$
For all neighborhoods $U$ of $I$, there exists a function $\delta : [0,1]\times [0,1] \to (0,\infty)$ such that for all finite covers $P$ of $[0,1]\times [0,1]$ by closed subrectangles with disjoint interiors, for all choice functions $t$ on $P$, if for all members $p$ of $P$, $P\subset \operatorname{B}(\delta(t(p)),t(p))$, then $\displaystyle\sum_{p\in P} \operatorname{area}(P)\cdot f(t(p)) \; \in \; U$.

Note that if we replaced $\delta : [0,1]\times [0,1] \to (0,\infty)$ with $\delta \in (0,\infty)$, we would get the Riemann integral. By proceeding as here but splitting into 4 subsquares instead of 2 subintervals, one can see that there will always be a a pair $P,t$ satisfying the relevant condition, and so $f$ has at most 1 gauge integral.

(ZFC)

Is it true that for all nonnegative functions $f : [0,1]\times [0,1] \to \mathbb{R}$,
1. if $f$ is gauge integrable, then $f$ is Lebesgue integrable
2. if $f$ is Lebesgue integrable, then $f$ is gauge integrable
3. if $f$ is integrable by both methods, then both integrals are equal
?

According to http://www.math.vanderbilt.edu/~schectex/ccc/gauge/, the analogues of 1,2, and 3 hold for the 1d gauge integral.

-
 This is another name for the Henstock-Kurzweil integral, right? (en.wikipedia.org/wiki/…) – Qiaochu Yuan Jan 21 2011 at 13:54 Yes. (filling in the char min) – Ricky Demer Jan 21 2011 at 23:24