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Let $f:[0,1] \to \mathbb{R}$ bounded,such that $\forall a $ with $0<a<1$, $f$ is integrable at the interval $ [a,1]$,show that f is integrable at $[0,1]$.

As $f$ is bounded, $\exists M>0$ such that $|f(x)| \leq M, \forall x \in [0,1]$.

Let $0<\epsilon<1$ .As $f$ is integrable at $[\epsilon,1]$, $\exists $ partition $P_1$ of $[\epsilon,1]$ ,such that:

$$U(f,P_1)-L(f,P_1)< \epsilon$$

Now,we consider the partition $P_2=\{0\} \cup P_1$ of $[0,1]$.

It is :

$$U(f,P_2)-L(f,P_2)=(\epsilon-0) \cdot \sup f([0, \epsilon])+U(f,P_1)-(\epsilon-0) \cdot \inf f([0, \epsilon])-L(f,P_1)$$

I jut wanted to know if $\sup f([0, \epsilon])=M \text{ and } \inf f([0, \epsilon])=-M$ or if it is just $\sup f([0, \epsilon]) \leq M \text{ and } \inf f([0, \epsilon]) \leq -M$.

Thanks in advance!

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    $\begingroup$ "Just $\sup f([0, \epsilon]) \leq M \text{ and } \inf f([0, \epsilon]) \leq -M$" (obviously). $\endgroup$ – Did Jun 24 '14 at 11:58
  • $\begingroup$ @Did I misunderstood. In my analysis class, a partition was always a family of intervals, not a set of division points. I see OP has it differently here, so I removed my comment. $\endgroup$ – 5xum Jun 24 '14 at 12:02
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    $\begingroup$ Edit to my previous comment: $\inf f([0, \epsilon]) \geq -M$. $\endgroup$ – Did Jun 24 '14 at 12:08
  • $\begingroup$ Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ – Martin Sleziak Jun 24 '14 at 13:11
  • $\begingroup$ I will remember it,for my next questions. $\endgroup$ – user159870 Jun 24 '14 at 13:12
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Choose $\delta<\text{mini}\{1,\frac{\epsilon}{2},\frac{\epsilon}{4M}\}$. And then choose a partition for the $\delta$. Then we will get $U(P_2,f)-L(P_2,f)\leq 2\delta M<\frac{\epsilon}{2}$ where $P_2=\{0\}$ and $P_1=\{\delta,\ldots,1\}$.

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  • $\begingroup$ Is it an other way to show what the exercise demands? $\endgroup$ – user159870 Jun 24 '14 at 12:35
  • $\begingroup$ No no. Just I filled the gap of the above argument with rigorous estimate. $\endgroup$ – Chellapillai Jun 25 '14 at 4:48

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