Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a homework question which is:

If $f(x)$ is a bounded function in $[0,1]$ and $\sup (f[\frac {1}{n},1])-\inf (f [\frac {1}{n},1])<\frac {1}{n}$ for every natural $n>0$. Prove that $f(x)$ is Integrable in $[0,1]$.

I have a feeling that this can be proven by showing somehow that $\inf(S_n-s_n)=0$ where $S_n$ and $s_n$ are the upper and lower limits of the split $P_n$.

However I have not managed to prove this, perhaps I am attacking this problem from the wrong direction.

Can someone help me out?

Thanks :)

share|improve this question
    
By bound, do you mean bounded? And by integrable, Riemann integrable? In this case fix $\varepsilon>0$, and $n$ such that $n^{-1}\leq\varepsilon$. Then construct step functions $s_1\leq fs_2$ such that $\int f-s_1\leq \varepsilon$ and $\int s_2-f\leq \varepsilon$. –  Davide Giraudo Jan 26 '12 at 9:48
    
@DavideGiraudo Yes I meant bounded but I did not mean Riemann integrable - We have not learnt Riemann's integral yet –  Jason Jan 26 '12 at 9:54
    
So do you mean Lebesgue integrable? –  Davide Giraudo Jan 26 '12 at 9:55
    
@DavideGiraudo I think the one I am familiar with is called Darboux integral - but I am not sure –  Jason Jan 26 '12 at 9:56
    
Ok, so do you know what you have to check? –  Davide Giraudo Jan 26 '12 at 10:01
show 1 more comment

2 Answers

up vote 1 down vote accepted

If you are familiar with the terminology of baby rudin (where the darboux integral is set up), the following reasoning might work. For every $n$ let $P_n=\{0,1/n,1\}$. Furthermore let $M_1=sup_{x\in[1/n,1]} f(x)$

and $M_0=sup_{x\in[0,1/n]} f(x)$ and similarly let $m_1=inf_{x\in[1/n,1]} f(x)$ and $m_0=inf_{x\in[0,1/n]} f(x)$

Now in the terminology of baby rudin we have:

$U(P_n,f)=(1/n)M_0+(1-1/n)M_1$ and $L(P_n,f)=(1/n)m_0+(1-1/n)m_1$

Let $\epsilon > 0$ and choose $n>2(M_0-m_0)/\epsilon$ and $(n-1)/n^2<\epsilon/2 ,$ then $U(P_n,f)-L(P_n,f)=(M_0-m_0)(1/n)+(M_1-m_1)(1-1/n)<\epsilon$

share|improve this answer
    
This seems like exactly what I'm looking for - I will give it a little better look in a minute, Thank you –  Jason Jan 26 '12 at 11:27
    
Yep this is exactly what I needed - Thank you very much. –  Jason Jan 26 '12 at 11:57
    
great, glad I could help. please, let me know if something is unclear. –  user22705 Jan 26 '12 at 12:00
    
Thanks everything is clear Maybe I should just add that I used $n>\max \left\{ \frac{2({{M}_{0}}-{{m}_{0}})}{\varepsilon },\frac{2}{\varepsilon },1 \right\}$ in my solution - which is pretty much the same thing just easier to type –  Jason Jan 26 '12 at 12:05
add comment

Fix $\varepsilon>0$ and $n$ such that $\max\left\{1,2\cdot\sup_{[0,1]}f\right\}n^{-1}\leq\varepsilon$. Consider the subdivision $0<n^{-1}<1$ and put $s(x)=\begin{cases}\inf_{0\leq t\leq n^{-1}}f(t)&\mbox{ if }0\leq x\leq n^{-1}\\\ \inf f\left([n^{-1},1]\right)&\mbox{ if }n^{-1}< x\leq 1,\end{cases}$ and $S(x)=\begin{cases}\sup_{0\leq t\leq n^{-1}}f(t)&\mbox{ if }0\leq x\leq n^{-1}\\\ \sup f\left([n^{-1},1]\right)&\mbox{ if }n^{-1}< x\leq 1,\end{cases}$

share|improve this answer
    
I Don't get why $n^-1 \le e$ –  Jason Jan 26 '12 at 10:49
    
Did you check if these functions work? The aim here is to take two step functions which are respectively greater/lower than $f$, and such the area between their graph is smaller than the fixed $\varepsilon$. –  Davide Giraudo Jan 26 '12 at 10:59
    
I am currently trying out a different method I think might work - I'll have a deeper look at these in a little bit :) Thanks –  Jason Jan 26 '12 at 11:01
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.