- #1
Demon117
- 165
- 1
I have been thinking about this for quite some time now. When I look at the function that descibes the fat cantor set namely:
f(x) = 1 for x[tex]\in[/tex]F and f(x) = 0 otherwise, where F is the fat cantor set.
I wonder, how do I prove that this is non-riemann integrable?
I have considered looking at the Riemann-Lebesgue theorem which gets me nowhere. So f is obviously bounded. But isn't this f discontinuous at all x[tex]\in[/tex][0,1]? This would imply that the discontinuity points of f need to be a zero set in order for it to be riemann integrable. But isn't the fat cantor set F not a zero set?
Any advice?
f(x) = 1 for x[tex]\in[/tex]F and f(x) = 0 otherwise, where F is the fat cantor set.
I wonder, how do I prove that this is non-riemann integrable?
I have considered looking at the Riemann-Lebesgue theorem which gets me nowhere. So f is obviously bounded. But isn't this f discontinuous at all x[tex]\in[/tex][0,1]? This would imply that the discontinuity points of f need to be a zero set in order for it to be riemann integrable. But isn't the fat cantor set F not a zero set?
Any advice?