Define function "f" on "A=\\{0\\}\\times[-1,1]\\cup[-1,1]\\times\\{0\\}" as "f(t)=0" for all "t\\in A" . Since "|f(t)|=1" for all "t\\in[-1,1]^2\\setminus A", we have that "|f(t)|\\le 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\\setminus A" , because "[-1,1]^2\\setminus A" is disjoint union of "(0,1]^2", "(0,1]\\times[-1,0)", "[-1,0)^2" and "[-1,0)\\times(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" .
Comments
Leave a comment