Let $I=[0,1]$ and $E \subset I$ be a finite subset of I. Let $f:I-> \mathbf{R}$ be a function on I such that $f(x)=0$ for any $x \not \in E$. Prove the $f$ is Riemann integrable on I without using Lebesgue integrability criterion and find $\int_I f$.

My attempt so far is;

Let $E=[a,b]$

let $\epsilon >0$ the there exists $\delta>0$

Then the Riemann sum of the interval I, $S(f;\dot{P})=S(f;\dot{P_1})+S(f;\dot{P_2})+S(f;\dot{P_3})$ where $\dot{P_1}$, $\dot{P_2}$ and $\dot{P_3}$ are any tagged partition in the intervals $[0,a]$, $[a,b]$ and $[b,1]$ respectively where their norms are less than $\delta$.

Since $f(x)=0 $ for all $x \not \in E$ we have that $S(f;\dot{P})=S(f;\dot{P_2})$.

The union of all subintervals in $\dot{P}$ with tags in [a,b] contains the interval $[a+\delta,b-\delta]$ and is contained within the interval $[b-\delta,a+\delta]$.

This is where I'm stuck. If I knew the function was constant I would say something like;

Since the value of the function is $f(a)$ at the start of the interval and $f(b)$ at the end, we have that $f(a)(a-b-2\delta)\leq S(f;\dot{P_2})\leq f(b)(a-b+2\delta)$

Where $f(a)=f(b)$ since it was constant, then I could rearrange it to have $|S(f;\dot{P_2})-f(b)(a-b)|\leq f(b)2\delta$ and just pick delta so it's less than epsilon.

But since I don't know the function I don't know what to do with possibly different values for $f(a)$ and $f(b)$.

  • $\begingroup$ What is your definition of Riemann integrability? $\endgroup$ – Pedro Tamaroff Dec 9 '13 at 5:00
  • $\begingroup$ A function $f:[1,b] \rightarrow$ if Riemann integrable on $[a,b]$ is $\exists L\in \mathbf{R}$ such that for every $\epsilon \geq 0$, $\exists \delta \geq 0$ such that if $\dot{P}$ is any tagged partition of $[a,b]$ with $||\dot{P}||\leq \delta$, then $|S(f;\dot{P})-L|\leq \epsilon$. $\endgroup$ – Duiliath Dec 9 '13 at 5:05
  • $\begingroup$ Can you solve the problem if the set E has exactly one element? The pf of the general case where E is finite is essentially same. $\endgroup$ – Raghav Dec 9 '13 at 5:11
  • $\begingroup$ If $E$ had only on element, wouldn't the length of the interval be zero? And if not wouldn't it just be the same as if the function was constant with $f(a)=f(b)$? $\endgroup$ – Duiliath Dec 9 '13 at 5:22

Denote $I=[0,1]$. Let $E=\{x_1,\dots,x_p\}$ be the set of points where $f(x)\neq 0$; so that $f=0$ on $[0,1]\setminus E$. Let $M=\max\limits_{1\leqslant i\leqslant p}|f(x_i)|$. We know $M>0$. Given $\varepsilon >0$, pick a partition $\Pi$ into intervals $I_1,\ldots,I_m$, and a set of tags $t_i\in I_i$, such that $\rVert \Pi\lVert<\frac{\varepsilon}{pM}$. Let $t_{i_j}$ denote the set of tags that coincide with some $x_j$. Note there are at most $p$ such tags, so in the following $l\leqslant p$. Then $$|R(\Pi,f)|=\left\lvert\sum_{j=1}^l f(t_{i_j})v(I_{i_j})\right\rvert\leqslant Mp \frac{\varepsilon}{Mp}=\varepsilon$$

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