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.
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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. 

