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


2dgauge$\displaystyle\int f \; = \; I$
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.


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, the analogues of 1,2, and 3 hold for the 1d gauge integral.

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This is another name for the Henstock-Kurzweil integral, right? ( – Qiaochu Yuan Jan 21 '11 at 13:54
Yes. (filling in the char min) – Ricky Demer Jan 21 '11 at 23:24

I'm no expert. However, a little googling indicates that there is a Fubini theorem for the gauge integral. So you can compute the 2d case by iterating 1d integrals. And in 1d, Lebesgue integrable implies gauge integrable, with equal values of the integrals, without any positivity hypotheses on f.

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