Define function f on A={0}×[−1,1]∪[−1,1]×{0} as f(t)=0 for all t∈A . Since ∣f(t)∣=1 for all t∈[−1,1]2∖A, we have that ∣f(t)∣≤1 on [−1,1]2, that is f is bounded on [−1,1]2.
Next, area of A equals 0, and f is continuous on [−1,1]2∖A , because [−1,1]2∖A is disjoint union of (0,1]2, (0,1]×[−1,0), [−1,0)2 and [−1,0)×(0,1] , and f is continuous on every of these sets.
Since [−1,1]2 is Jordan mesurable, f is bounded and continuous almost everywhere on [−1,1]2, by Lebesgue's criterion for Riemann integrability we obtain that f is integrable on [−1,1]2 .
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