# Proving a bounded function is integrable

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 :)

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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

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

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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. – Henrik 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

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 $S(x)=\begin{cases}\sup_{0\leq t\leq n^{-1}}f(t)&\mbox{ if }0\leq x\leq n^{-1}\\\ \sup f\left([n^{-1},1]\right)&\mbox{ if }n^{-1}< x\leq 1,\end{cases}$

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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