Shouldn't the first integral have (b-a) rather than 1 on the end? (i.e the width of the rectangle, the height always being 1). And why is there an inequality sign on the end? Why not just equal?
Thank you!
Darboux integral mistake in example?
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analysis
riemann-sum
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0$b-a=1$ I think that explains first part. – 2017-02-11
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0Thanks, Still don't understand why it's 1? surely the width of the rectangle would be (b-a) and the height 1, so the area is 1 times (b-a)? – 2017-02-11
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0It says (a,b) is a subset of [0,1] though? I thought that meant (a,b) could be any interval in between? – 2017-02-11
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0Oops, sorry :). I don't know why we must switch to subintervals. The STANDARD argument may go like that: once you have a step function $\varphi \le$ the Dirichlet function everywhere on $[0,1]$ it must be the zero function on $[0,1].$ Next, if a step function $\varphi$ majorizes the Dirichlet function on $[0,1],$ it must be the constant function taking the value $1.$ In effect, the lower Darboux integral over $[0,1]$ is zero, and the upper Darboux integral over $[0,1]$ is one. – 2017-02-11
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0Both statements on the step functions above use the fact that that any open interval $(a,b)$ does indeed contain a rational point. It seems the author of the textbook simply forgot to replace $a$ with $0$ and $b$ with $1,$ when they obtained the Darboux integrals. – 2017-02-11