shegiggles
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Let f: [a,b] -> R. Given f is integrable and bounded below, show 1/f is integrable.
Analysis: Given f is integrable and bounded below, show that
- Thread starter shegiggles
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Let f(x)=x on [0,1]. It's bounded below and integrable. However, 1/f defined on (0,1] is not integrable on its domain!
Perhaps you want to prove this for functions f defined on [a,b] for which 1/f is also defined on [a,b]? Also, are we talking about Riemann integral or Lebesgue integral?? Please be specific and more importantly, show how far YOU got with the problem.
Perhaps you want to prove this for functions f defined on [a,b] for which 1/f is also defined on [a,b]? Also, are we talking about Riemann integral or Lebesgue integral?? Please be specific and more importantly, show how far YOU got with the problem.
shegiggles
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mark07 said:
Thanks for the reply.
Its about Riemann integral.
I wish I can show you how far I got with the problem but I dont even know how to start this. Am so lost. Please help me.
Thanks
pka
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There are huge difficulties in trying to help with a question on this level. Chief among them is the matter of definitions. There are many ways that different textbooks define the Riemann Integral. We do not know which one you are using. So the best I can do is give you a broad outline.
But first, as Mark has shown you that as stated the proposition is false even in his modified form, i.e. 1/f is defined in [a,b]. This problem is most often stated that there is a c>0 and f(x)>c for all x in [a,b]. With that additional condition then 1/f is integrable in [a,b].
Here an outline of a proof. If \(\displaystyle [u,v] \subseteq [a,b]\) and \(\displaystyle M(f,[u,v])\) is the maximum of \(\displaystyle f\) on \(\displaystyle [u,v]\) and \(\displaystyle m(f,[u,v])\) is the minimum \(\displaystyle f\) on \(\displaystyle [u,v]\) we can see that \(\displaystyle M\left( {\frac{1}{f},[u,v]} \right) = m(f,[u,v])\quad \& \quad m\left( {\frac{1}{f},[u,v]} \right) = M(f,[u,v]).\)
Now because \(\displaystyle f\) is integrable on \(\displaystyle [a,b]\) use the upper and lower sums to complete the problem.
Again, the way suggested may not fit your definitions.
But first, as Mark has shown you that as stated the proposition is false even in his modified form, i.e. 1/f is defined in [a,b]. This problem is most often stated that there is a c>0 and f(x)>c for all x in [a,b]. With that additional condition then 1/f is integrable in [a,b].
Here an outline of a proof. If \(\displaystyle [u,v] \subseteq [a,b]\) and \(\displaystyle M(f,[u,v])\) is the maximum of \(\displaystyle f\) on \(\displaystyle [u,v]\) and \(\displaystyle m(f,[u,v])\) is the minimum \(\displaystyle f\) on \(\displaystyle [u,v]\) we can see that \(\displaystyle M\left( {\frac{1}{f},[u,v]} \right) = m(f,[u,v])\quad \& \quad m\left( {\frac{1}{f},[u,v]} \right) = M(f,[u,v]).\)
Now because \(\displaystyle f\) is integrable on \(\displaystyle [a,b]\) use the upper and lower sums to complete the problem.
Again, the way suggested may not fit your definitions.