Question #85699
Show that the function f:[2,3]→R defined by :
f(x)={0 if x is rational {1 if x is irrational
is discontinues and not integrable over [2,3].Does it imply that every discontinues function is non-integrable?Justify your answer.
1
Expert's answer
2019-03-02T16:29:41-0500

Since Q\mathbb{Q} is dense in R\mathbb{R}, we have that supx[xi,xi+1]f(x)=1\sup_{x\in [x_i, x_{i+1}]}f(x)=1 and infx[xi,xi+1]f(x)=0\inf_{x\in [x_i, x_{i+1}]}f(x)=0 on any interval [xi,xi+1]R[x_i, x_{i+1}] \subset \mathbb{R}, therefore the upper and lower sums don't converge to the same limit. Hence ff is not Riemann-integrable.


It does not imply that every discontinues function is non-integrable because g(x)=xg(x)=x for x[0,1)x\in [0,1) and g(x)=x+1g(x)=x+1 for x[1,2]x\in [1,2] is integrable on [0,2][0,2] and have discontinuity at x=1x=1.


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